Static Electricity Review

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Part A: Multiple Choice

1. Which of the following are true of static charges? Choose all that apply.

  1. Like charges repel.
  2. Like charges attract.
  3. Opposite charges repel.
  4. Opposite charges attract.
  5. A positively charged object has lost electrons.
  6. A positively charged object has gained protons.
  7. A negatively charged object has lost protons.
  8. A negatively charged object has gained electrons.

 

Answer: ADEH

A and D are true; they are the simple statement of our essential charge interactions, best remembered by the jingle: "opposites attract, likes repel."

B and C are false for these same reasons; they violate the basic statement of charge interactions.

E and H are true. For an object to become charged, it must either gain or lose electrons. Losing electrons results in more positive charge than negative charge, making the object charged positively. Gaining electrons results in more negative charge than positive charge, making the object charged negatively. Protons are tightly bound in the nucleus of atoms and can never be added nor removed from atoms by ordinary electrostatic methods. The same reasoning leads one to reject choice F and G as possible true statements; the suggest that protons can be added or removed.



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The Structure of Matter || Neutral vs. Charged Objects || Charge Interactions

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2. Identify the following objects as being either

a. positive

b. negative

c. neutral

If there is no conclusive evidence, then select all that could be true.


Description of Object

a, b, or c?

i. An object possesses more protons than electrons.

A

ii. An object possesses more neutrons than electrons.

ABC

iii. A formerly neutral object that just lost some electrons.

A

iv. A formerly neutral object that just gained some electrons.

B

v. An object which attracts a negatively-charged balloon.

AC

vi. An object which attracts neutral paper bits and attracts a negatively-charged balloon.

A

vii. An object which attracts neutral attracts paper bits and repels a negatively-charged balloon.

B

viii. An object which attracts a negatively-charged balloon and attracts a positively-charged balloon.

C

ix. An object which attracts a charged balloon (balloon A) which is attracted to a negatively-charged balloon (balloon B).

ABC

x. An object which attracts a balloon (balloon C) which is repelled by a negatively-charged balloon (balloon D).

AC

xi. An object which repels a balloon (balloon E) which is repelled by a positively-charged balloon (balloon F).

A

xii. An object around which the electric field vector is directed inwards.

B

xiii. An object around which the electric field vector is directed outwards.

A

Answer: See table above

i. Protons are positive; electrons are negative; more protons than electrons would mean an overall positive charge.

ii. Neutrons are neutral and will not have an influence on the overall charge. So this object could be +, -, or neutral depending on the relative number of protons and electrons.

iii. If a neutral object loses some electrons, then it will possess more protons (positive charge) than electrons (negative charge); the object will have an overall + charge.

iv. If a neutral object gains some electrons, then it will possess more electrons (negative charge) than protons (positive charge); the object will have an overall - charge.

v. A negatively charged object would never be attracted to a - balloon. However, a + object would be attracted to a - balloon (opposites attract) and a neutral object would be attracted to a - balloon (since neutral objects are attracted to any charged object).

vi. An object which attracts neutral paper bits must be charged. If the object also attracts a - balloon, then it must have a + charge (opposites attract).

vii. If an object has a repulsive interaction with any other object, then it is definitely charged. Since like charges repel, the charge which it has is the same type of charge as the charge on the object which it repels. This object has a - charge.

viii. A neutral object will be attracted to any charged object, whether + or -. In this case, the object attracts both + balloons and - balloons. Only a neutral object could do this.

ix. Balloon A is either + (since it attracts a - object) or neutral (since neutral objects will also attract any charged object). If our object attracts balloon A, then it is either + or - (since such objects would attract balloon A if it were neutral) or even neutral (since it would attract balloon A if balloon A were +).

x. Balloon C must be - since it repels another - balloon (like charges repel). If our object attracts balloon C, then it is either + (since opposites attracts) or neutral (since neutral objects are attracted to any charged object).

xi. Balloon E must be + if it repels balloon F (like charges repel). If our object repels balloon E, then it to must have a + charge.

xii. By definition, the electric field vector points in the direction which a + test charge would be accelerated. So by extension, the E vector points towards - charges and away from + charges. This object must have a - charge.

xiii. Using the same reasoning as above, this object must have a + charge.



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The Structure of Matter || Neutral vs. Charged Objects || Charge Interactions

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3. Consider the following statements given below and determine if the charge of Object A is:

a. positive

b. negative

c. neutral

Description of Charging Method

a, b, or c?

i. Object A is charged by friction using animal fur. Animal fur has a greater electron affinity than object A.

A

ii. Object A is charged by contact using a negatively charged object.

B

iii. Object A is charged by induction using a positively charged object.

B

iv. Object A is used to charge an aluminum plate by the method of induction. The aluminum plate acquires a positive charge.

B
v. Object A is used to charge an aluminum plate by the method of induction. The aluminum plate acquires a negative charge.

A

vi. A rubber rod has a greater electron affinity than animal fur. The rubber rod is charged by friction with animal fur. The rubber rod is then used to charge Object A by the method of contact.

B

vii. A rubber rod has a greater electron affinity than animal fur. The rubber rod is charged by friction with animal fur. The rubber rod is then used to charge Object A by the method of induction.

A

viii. A rubber rod has a greater electron affinity than animal fur. The rubber rod is charged by friction with animal fur. The rubber rod is then used to charge an aluminum pop can by the process of induction. The pop can is then contacted to object A.

A

Answer: See above table.

i. When two objects are charged by friction, electrons are transferred between objects. The object made of a material with the greatest electron affinity is the object which receives the electrons; the other object loses electrons. Thus, Object A loses electrons and acquires a + charge.

ii. When charging an object by contact, the object receives the same type of charge as the object used to charge it.

iii. When charging an object by induction, the object receives the opposite type of charge as the object used to charge it. The answer to question 12 includes a detailed explanation of the induction charging method.

iv. Since induction charging gives objects opposite types of charge, Object A must have an opposite charge as the aluminum plate.

v. The same reasoning used in iv. above can be used to explain this question.

vi. The rubber rod must become charged negatively since it would acquire electrons from the animal fur. It would then charge Object A with the same type of charge (-) since contact charging results in charging an object with the same type of charge.

vii. The rubber rod must become charged negatively since it would acquire electrons from the animal fur. It would then charge Object A with the opposite type of charge (+) since induction charging results in charging an object with the opposite type of charge.

viii. The rubber rod must become charged negatively since it would acquire electrons from the animal fur. It would then charge the aluminum plate with the opposite type of charge (+) since induction charging results in charging an object with the opposite type of charge. The aluminum plate would then charge Object A positively since contact charging results in charging an object with the same type of charge.


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Charging by Friction || Charging by Induction || Charging by Conduction

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4. A neutral plastic strip is rubbed with cotton and acquires a positive charge. Which of the following statements are true of the positively-charged strip?

  1. It lost some electrons to the cotton during the charging process.
  2. It lost all of its electrons to the cotton during the charging process.
  3. It has the opposite charge as the cotton.
  4. It would now be repelled by the piece of cotton which was used to charge it.
  5. It gained protons during the rubbing process.
  6. As a material, plastic has a greater affinity for electrons than cotton.
  7. It could exert either a repulsive or attractive influence upon neutral paper bits.
  8. It has an excess of protons compared to the number of electrons.
  9. It could be used to charge an electroscope negatively by the process of induction.
  10. It lost negative electrons and gained positive electrons during the charging process.
  11. It lost neutrons during the charging process (or at the very least, its neutrons became ineffective).


Answer: ACHI

a. During charging by friction, electrons are transferred from one object to the other object. The object which acquires a + charge is the object which loses the electrons.

b. While some electrons are lost, not all electrons lost.

c. Charging by friction results in two objects with the opposite type of charge.

d. If the two rubbed objects are brought near, then they will attract (rather than repel) since they are charged oppositely.

e. Protons are never gained or lost during ordinary electrostatic experiments. They are tightly bound in the nuclei of atoms; it would require an atom-smasher to induce protons to move.

f. The object which has the greater electron affinity is the one which acquires the - charge during the friction charging process.

g. A neutral object like paper could only attract a charged object like the plastic; repulsion is only observed of two charged objects having the same type of charge.

h. Since it has lost electrons, the plastic will have an excess of protons.

i. If a positively-charged object is used to charge an electroscope by induction, then the electroscope would acquire a - charge. Induction charging results in objects with opposite types of charge.

j. There is no such thing as positive electrons (at least not for our purposes).

k. As you may have noticed, talk of neutrons in a unit on electrostatics is largely absent. Neutrons have nothing to do with the discussion.



Useful Web Links
Charging by Friction || Charging by Induction || Charging by Conduction

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5. A positively-charged glass rod is touched to the plate of a neutral electroscope. Upon contact, the electroscope becomes charged and the needle deflects. Which of the following statements are true of the charged electroscope?

  1. The electroscope is now charged positively.
  2. The electroscope and the glass rod now have the same type of charge.
  3. The electroscope was charged by the method of induction.
  4. The electroscope gained protons during the charging process.
  5. The electroscope gained electrons during the charging process.
  6. The electroscope lost all of its electrons during the charging process.
  7. During the charging process, some electrons left the electroscope and entered the glass rod.
  8. The number of electrons present in the electroscope is less than the number of protons.
  9. The electroscope needle will deflect even more if the glass rod is brought near it again.
  10. The electroscope needle would slowly approach the neutral position if a negatively-charged balloon is brought near.

Answer: ABGHIJ

a. Charging by contact places the same type of charge on the object being charged. So if a positively charged rod is used to charge the electroscope, the electroscope becomes charged positively.

b. Same as part a.

c. Induction charging would involve bringing the glass rod near to the electroscope without making contact. The method described above is the contact charging method.

d. Protons are never gained or lost during ordinary electrostatic experiments. They are tightly bound in the nuclei of atoms; it would require an atom-smasher to induce protons to move.

e. The glass rod, being charged positively, would attract electrons which were present upon the electroscope. These electrons would be transferred to the glass rod. The electroscope would lose electrons, not gain them. In fact, that is why it becomes charged positively.

f. While the electroscope does lose electrons, it does not lose all its electrons.

g. See the explanation in part e above.

h. Having lost electrons, there are now more protons in the electroscope than there are electrons.

i. The electroscope and the glass rod are both positively-charged. The presence of the glass rod near the positively-charged electroscope would induce the needle to deflect even more rather than cause it to move towards the neutral position.

j. Since the electroscope is now charged positively, bringing a negatively charged object nearby would cause the needle to return to the neutral position.



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Charging by Conduction

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6. A negatively-charged balloon is brought near to (without touching) a neutral electroscope. With the negatively-charged balloon held near, the electroscope is momentarily touched by a  hand (ground). The balloon is then removed and the needle deflects, thus showing a charge. Which of the following statements are true of the charged electroscope?

  1. The electroscope was charged by the method of induction.
  2. The charge on the electroscope is the same type of the charge on the balloon.
  3. The electroscope has more protons than electrons.
  4. The electroscope gained protons during the charging process.
  5. The electroscope gained electrons during the charging process.
  6. The electroscope lost all of its electrons during the charging process.
  7. The electroscope lost some electrons during the charging process.
  8. During the charging process, protons moved from the electroscope to the balloon.
  9. During the charging process, electrons moved from the electroscope to the balloon.
  10. During the charging process, electrons moved from the electroscope to the hand (ground).
  11. During the charging process, electrons moved from the hand (ground) to the electroscope.

 

Answer: ACGJ

a. This is a perfect description of the induction charging method.

b. Induction charging places a charge on an object which is opposite of the type of charge on the object used to charge it. Thus, the electroscope acquires a positive charge.

c. Since the electroscope is positively charged, there are more protons than electrons on the electroscope.

d. Protons are never gained or lost during ordinary electrostatic experiments. They are tightly bound in the nuclei of atoms; it would require an atom-smasher to induce protons to move.

e. When the - balloon was brought near, electrons in the electroscope move to a location far from the balloon (like-charged objects repel). When touched by the hand (ground), the electrons move even further, entering the hand and the person. So in the induction charging process, the electroscope lost electrons.

f. While the electroscope loses electrons, it does not lose all of its electrons.

g. See explanation to step e above.

h. Protons are never gained or lost during ordinary electrostatic experiments. They are tightly bound in the nuclei of atoms; it would require an atom-smasher to induce protons to move.

i. Electrons cannot move from the balloon to the electroscope because there is no available conducting pathway. The air separates the two objects and serves as an insulator, thus preventing the movement of electrons between them.

j. See explanation to step e above.

k. See explanation to step e above.



Useful Web Links
Charging by Induction

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7. The outer shell electrons in metals are not tightly bound to the nuclei of their atoms. They are free to roam throughout the material, moving from atom to atom. These materials are good ____.

  1. conductors
  2. insulators
  3. for nothing

Answer: A

Metals are great conductors. That is, electrons are free to move across the surface of metals, migrating from atoms to atoms. Being conductors, metals can serve as pathways for the movement of electrons from atom to atom, molecule to molecule, and object to object. All of this is due to the fact that the outer electrons (also known in some chemistry circles as valence electrons) are not tightly bound to the nuclei of their atoms.



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Conductors and Insulators

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For Questions #8 and #9, consider the following situation. Connor Duct (Con to his friends) takes a positively charged rubber rod and touches a metal sphere on an insulated stand as shown at the right. Draw the direction of electron flow.

8. The charge on the metal sphere will end up ___.

  1. neutral
  2. negative
  3. positive

 

Answer: C

This is an example of charging by contact. When a neutral object is charged by contact, the object acquires the same type of charge as the object used to charge it. In this case, the sphere acquires the same type of charge as the rubber rod - a positive charge. (As for how this happens, see the next question.)



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Charging by Conduction

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9. The sphere acquires this charge because ____.

  1. electrons move from the rubber rod to the sphere
  2. electrons move from the sphere to the rubber rod
  3. protons move from the rubber rod to the sphere
  4. protons move from the sphere to the rubber rod
  5. the rubber rod creates a charge on the sphere

Answer: B

Once touched, the sphere and the rubber rod transfer electrons between them. Since the rubber rod is + and electrons are -, the rubber rod attracts electrons from the sphere towards itself. Some of these electrons move into the rubber rod, thus reducing the amount of + charge upon it. The sphere in turn becomes charged positively since it has lost electrons.



Useful Web Links
Charging by Conduction

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10. If a positively charged plate is brought near the top of a positively-charged electroscope, then the deflected needle will ____.

  1. not move at all
  2. be deflected more
  3. be deflected less

 

Answer: B

A positively charged electroscope has protons and electrons; only there would be more protons compared to the number of electrons. There is an excess of + charge ("protons without partners") uniformly distributed over the electroscope. Bringing another + object nearby, will draw electrons out of the needle of the electroscope up into the plate. This would leave and even greater excess of charge in the needle of the electroscope. As such, the needle would be seen to deflect even more.



Useful Web Links
Charge Interactions || Charging by Induction

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11. If an electroscope, charged up with excess negative charge, is touched and grounded, then it will end up with ___ charge.

  1. no
  2. a negative
  3. a positive

Answer: A

When a charged object is grounded, there is a movement of electrons between the object and the ground until the object has a balance of charge. In this case, electrons move from the electroscope to the ground until the number of electrons equals the number of protons.



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Grounding - the Removal of a Charge

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12. A negatively charged balloon is brought near a metal can that rests on a wood table. The side of the can opposite the balloon is momentarily touched. The can is then _______.

  1. positively charged
  2. negatively charged
  3. uncharged

Answer: A

This is an example of charging by induction. In all such cases, the object being charged acquires a charge that is opposite in type to the object used to charge it. Bringing the - balloon near the can will chase electrons in the can to the far side of the can. When touched on the far side, electrons are chased further, exiting the can and migrating into the finger and the person. Having lost electrons, the can now has an excess of positive charge. In this example, the finger and person serve as the ground - a seemingly infinite reservoir or holding place of charge.



Useful Web Links
Charging by Induction

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13. Charge carriers in a metal are electrons rather than protons. This is due to the fact that electrons are __________.

  1. loosely bound
  2. lighter
  3. far from a nucleus
  4. all of the above
  5. none of the above

Answer: A

Electrons are light and they are far from the nucleus, but neither of these features explain why they are the carriers of charge in electrostatic experiments. Electrons, unlike the protons, are not bound up in an inescapable condition within the atoms of metals. As such, they can move from atom to atom throughout the surface of the metal and can even from object to object. On the other hand, protons are tightly bound in the nucleus of an atom and are unable to escape by ordinary electrostatic means.



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The Structure of Matter || Conductors and Insulators
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14. Two like charges ________.

  1. attract each other
  2. repel each other
  3. neutralize each other
  4. have no effect on each other
  5. must be neutrons

Answer: B

There are three charge interactions which should be known; this is one of them: Like charged objects repel. The other two: i) oppositely-charged objects attract, and ii) a neutral object and any charged object (whether + or -) will attract.



Useful Web Links
Charge Interactions

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15. If you comb your hair and the comb becomes positively charged, then your hair becomes _______.

  1. positively charged
  2. negatively charged
  3. uncharged

Answer: B

This is an example of charging by friction. Two neutral objects, when rubbed together, will transfer electrons between them. The object with the greatest electron affinity will acquire electrons from the other object; the two become charged oppositely. In this case, the hair acquires electrons from the comb, leaving the comb with a shortage of electrons and a + charge; the hair now has an excess of electrons and a - charge.



Useful Web Links
Charging by Friction

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16. Which of the following charging methods can result in charging an object with a negative charge? Choose all that apply.

a. charging by friction

b. charging by contact

c. charging by induction


Answer: ABC

If object A is rubbed with another object having a lower electron affinity, then the atoms of object A will steal electrons from the other object and acquire a negative charge.

If object A is touched by a negatively-charged object, then it will become charged negatively.

If a positive charge object is used to charge object A by the method of induction, then object A will acquire a negative charge as it draws electrons from the ground.



Useful Web Links
Charging by Friction || Charging by Induction || Charging by Conduction

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17. Which of the following charging methods work without ever touching the object to be charged to the object used to charge it? Choose all that apply.

a. charging by friction

b. charging by contact

c. charging by induction



Answer: C

In charging by induction, a charged object is brought near an object without touching it. The presence of the charge object induces electron movement and a polarization of the object. Then conducting pathway to ground is established and electron movement occurs between the object and the ground. During the process, the charged object is never touched to the object being charged.



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Charging by Friction || Charging by Induction || Charging by Conduction

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18. Which of the following charging methods result in charging an object opposite to the charge on the object used to charge it? Choose all that apply.

a. charging by friction

b. charging by contact

c. charging by induction



Answer: AC

In charging by friction, the two objects being rubbed acquire opposite types of charge as electrons are transferred between objects.

In charging by contact, both objects acquire the same type of charge.

In charging by induction (as described in the answer to Question #17), the objects acquire the opposite type of charge.



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19. Which of the following statements are true statements about electric force? Choose all that apply.  

  1. Electric force is a contact force.
  2. Electric forces can only act between charged objects - either like-charged or oppositely-charged.
  3. Electric forces between two charged objects increases with increasing separation distance.
  4. Electric forces between two charged objects increases with increasing quantity of charge on the objects.
  5. If object A attracts object B with an electric force, then the attractive force must be mutual - i.e., object B also attracts object A with the same force.
  6. A doubling of the quantity of charge on one of the objects results in a doubling of the electric force.
  7. A tripling of the quantity of charge on both of the objects results in an increase in the electric force by a factor of 6.
  8. A doubling of the separation distance between two point charges results in a quadrupling of the electric force.
  9. A tripling of the separation distance between two point charges results in an electric force which is one-sixth of the original value.

 

Answer: DEF

a. Electric force is a non-contact force (or field force); it can act over separation distances even when the objects do not touch.

b. An electrical attraction can even occur between a charged object and a neutral object. The neutral object is first polarized and then the attraction can occur.

c. As separation distance increases, the electric force decreases; there is an inverse relationship.

d. As the charge on any of the two objects increases, the electric force increases; there is a direct relationship.

e. Forces - even electric forces - come in pairs. For every action, there is an equal and opposite reaction force. Indeed, to say that "object A attracts object B" is to automatically infer that object B also attracts object A.

f. Whatever alteration is in made of the charge, the same alteration is made of the electric force. Doubling the charge will double the force.

g. Whatever alteration is in made of the charge, the same alteration is made of the electric force. Tripling the charge will triple the force; tripling the other charge will triple the force again. Tripling the force (x3) two times will have the net effect of increasing the force by a factor of 9 (not 6).

h. Force is inversely proportional to the square of the distance. Whatever alteration is made of the distance, the inverse square alteration is made of the force. So increasing distance by a factor of 2 (doubling it) will decrease force by a factor of 4.

i. Force is inversely proportional to the square of the distance. Whatever alteration is made of the distance, the inverse square alteration is made of the force. So increasing distance by a factor of 2 (doubling it) will decrease force by a factor of 9 (3^2).



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20. Which of the following statements are true statements about electric field? Choose all that apply.

  1. The electric field strength created by object A is dependent upon the separation distance from object A.
  2. The electric field strength created by object A is dependent upon the charge on object A.
  3. The electric field strength created by object A is dependent upon the charge of the test object used to measure the strength of the field.
  4. The electric field strength about charged Object A is the force per charge experienced by a test charge placed at some location about Object A.
  5. As the distance from Object A is doubled, the electric field strength created by object A increases by a factor of 4.
  6. As the charge of Object A is doubled, the electric field strength created by object A increases by a factor of 2.
  7. As the charge of Object A is doubled and the distance from object A is doubled, the electric field strength created by object A decreases by a factor of 2.
  8. Object B is used to test the electric field strength about Object A; as the charge of Object B is doubled, the force which it experiences is doubled but the electric field strength remains the same.
  9. Object B is used to test the electric field strength about Object A; as the separation distance between Object A and Object B is doubled, the force which it experiences decreases by a factor of 4 but the electric field strength remains the same.
  10. The electric field strength inside of a closed conducting object (e.g., inside the sphere of the Van de Graaff generator) is zero.
  11. For irregularly shaped objects, the electric field strength is greatest around points of highest curvature.
  12. The electric field is a vector which points in the direction which a positive test charge would be accelerated.
  13. Electric fields are directed inwards in regions around negatively-charged objects and outwards in regions around positively-charged objects.
  14. Units on electric field are Newtons/Coulomb (N/C). 

Answer: ABDFGHJKLMN

a. TRUE The E vector depends upon two factors - the charge of the object creating the field and the separation distance from that charge.

b. TRUE The E vector depends upon two factors - the charge of the object creating the field and the separation distance from that charge.

c. FALSE An object can be used to measure the strength of an electric field; such an object is called a test charge. The test charge encounters an electric force. The electric field is the ratio of the force divided by the charge on the test charge. If the quantity of charge on the test charge is increased, the force is increased proportionately but the ratio remains the same.

d. TRUE An object can be used to measure the strength of an electric field; such an object is called a test charge. The test charge encounters an electric force. The electric field is the ratio of the force divided by the charge on the test charge.

e. FALSE The electric field strength is inversely proportional to the square of the distance of separation from the charge. If the distance is increased by a factor of 2, then the E is decreased by a factor of 4.

f. TRUE The electric field strength is directly proportional to the quantity of charge on the object which creates the field. Doubling the charge (Q) will double the electric field strength.

g. TRUE The electric field strength is directly proportional to the quantity of charge on the object which creates the field and inversely proportional to the square of the distance of separation from that charge. Doubling the charge will have the effect of increasing the E by a factor of 2; doubling the distance will have the effect of decreasing the E by a factor of 4. The combined effect of these two factors will result in decreasing the E by a factor of 2.

h. TRUE The electric field strength (E) created by object A does not depend upon the charge on the object used to test it. The electric field force (F) would be doubled by the doubling of the charge. But the electric field strength (E) is the force per charge - twice the force on twice the charge yields the same electric field strength.

i. FALSE The electric field strength created by object A does depend on how far from object A that the test charge is. Twice the separation distance means one-fourth the electric force (the inverse square law) and therefore one-fourth of the electric field.

j. TRUE This is a unique property of the electric field inside a closed conductor.

k. TRUE This is another unique property of the electric field. This property explains how lightning rods work. Being pointed, the electric field strength is large around the points (parts of highest curvature).

l. TRUE This is the customary convention used to define the direction of the electric field around any charged object.

m. TRUE Because electric field direction is defined as the direction which a positively-charged object would accelerate, the direction would be toward negatively charged objects (+ test charges are attracted) and away from positively-charged objects (+ test charges are repelled).

n. TRUE Since electric field is the force per charge exerted upon a test charge, the units would be force units per charge units - Newtons/Coulomb.



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21. Which of the following statements are true statements about lightning rods? Choose all that apply.

  1. Lightning rods are placed on homes to reduce the risk of lightning damage to a home.
  2. Any metallic object which is placed on the roof of a home and grounded by an appropriate conducting pathway can serve as a lightning rod.
  3. Most lightning rods are pointed as a decorative feature.
  4. To be totally effective, a lightning rod must stretch high into the sky and draw charge from the lowest clouds by the method of contact.
  5. Lightning rods are capable of reducing the excess charge buildup in clouds, a characteristic of dangerous thunderstorms.
  6. Charge is incapable of passing from clouds to a lightning rods since the air between the clouds and the lightning rod has an insulating effect.
  7. The electric field strength about the points of lightning rods are very high.

Answer: AG

a. TRUE The role of lightning rods is to divert charge around your home to the ground in the event of a possible lightning strike.

b. FALSE If the lightning rod does not have the characteristic protrusion up into the air above the building it protects, then it might fail to do what it does best - divert charge around the building and preventing it from being conducted through the home.

c. FALSE Lightning rods are pointed because the electric field around a sharply curved object is high and serves to ionize air around it; this provides a conducting pathway from the cloud to the ground. The pathway allows charge to slowly flow between cloud and ground, thus preventing a sudden lightning discharge. In more recent years, scientists have found that a blunt tip on the lightning round provides and equal if not greater measure of protection.

d. FALSE Lightning rods do not need to contact the clouds to discharge them. The strong electric fields around the points serves to turn air from an insulator to a conductor (by ionizing particles in the air). This provides a conducting pathway between the cloud and the lightning rod.

e. FALSE This statement is a good description of the classical dissipation theory of lightning rods.  It was once believed (beginning with Ben Frnaklin) that lightning rods served to dissipate or remove electrostatic charge buildup in clouds by removing the charge gradually over time.  Scientists generally agree that that dissipation theory is not an accurate view of the role of lightning rods.  It is believed that lightning rods serve to divert electrostatic charge around a home during a lightning strike.

f. FALSE If this were true, then there wouldn't be any lightning strikes.  The role of the lightning rod is to actually make the pathway between the cloud and the ground more conductive by ionizing the air around the rod and providing a more conductive path from cloud to ground through the lightning rod.

g. TRUE This explains why lightning rods are often pointed. Though as discussed in more detail in c. above, blunt-tipped lighning rods have been found to provide equal or even greater protection as pointed-tip lightning rods.



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22. Charged balloons are used to induce a charge upon neutral metal pop cans. Identify the type of excess charge which would be present on cans G, H, I, J, K, L, M, N and P as shown in the diagrams below.

a. positive

b. negative

c. neutral

 
Diagram
Type of Charge on...

I.

 

G: positive (A)

H: negative (B)

II.

 

I: negative (B)

J: positive (A)

II.

 

K: positive (A)

L: negative (B)

M: positive (A)

IV.

 

N: negative (B)
V.
 

P: neutral (C)

Answer: See table above.

I. The presence of the negatively-charged balloon repels electrons which are present in can G. Many of these electrons move into can H. This leaves can G charged positively (it lost electrons) and can H charged negatively (it gained electrons).

II. The presence of the negatively-charged balloon next to can J repels electrons which are present in can J. Many of these electrons move into can I. Similarly, the presence of the positively-charged balloon next to can I will attract electrons present in both cans. This combined effect leaves can J to become charged positively (it lost electrons) and can I to become charged negatively (it gained electrons).

III. The presence of negatively-charged balloons on each end of the collection of cans causes electrons to be repelled from the balloons. Many electrons leave the two cans on the ends and enter the can in the middle in order to distant themselves from the balloons. This leaves Can L charged negatively (it gained electrons) and Cans K and M charged positively (they lost electrons).

IV. The presence of the positively-charged balloon will attract electrons towards itself. There is a migration of electrons from the ground (hand) towards the balloon. This leaves can N charged negatively.

V. Electrons in can P will be attracted to the positively-charged balloon. This will polarize the can. But unless there is a source of electrons connected to can P, there is no way for the can to develop an overall charge. The can remains neutral.



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23. In diagram IV of Question #22, the pop can acquires the charge that it does because ___.

  1. electrons move from the balloon to the pop can
  2. electrons move from the pop can to the balloon
  3. protons move from the balloon to the pop can
  4. protons move from the pop can to the balloon
  5. electrons move from the pop can to the hand (ground)
  6. electrons move from the hand (ground) to the pop can
  7. protons move from the pop can to the hand (ground)
  8. protons move from the hand (ground) to the pop can
 

Answer: F

Any explanation which involves the movement of protons can be quickly ruled out since protons are bound in the nucleus and incapable of moving about during ordinary electrostatic experiments. Electrons cannot move from the balloon to the pop can (nor vice versa) since there is no connecting path between these two objects. Air is a relatively good insulator and prevents the movement of charge between these two objects. What happens is that electrons in the hand sense the electric pull of the balloon and are drawn towards it, entering the pop can and imparting a negative charge to it.



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Charging by Induction

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24. A negatively charged balloon will be attracted to a neutral wooden cabinet due to polarization. Which one of the following diagrams best depict why this occurs? 

 

Answer: D

The molecules of the wooden cabinet are neutral (which rules out diagrams A, B and E) yet polarized. The molecules polarize in such a way so as to make its forward side most appealing to the balloon. This means that the positive poles of the wood molecules are nearest the negative charge of the balloon. Only diagram D shows this orientation.



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Polarization

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For Questions #25 - #27, identify the type of charge on objects A-D based on the electric field lines shown for each configuration of charges.

a. positive

b. negative

c. neutral

25.



A is positive

26.



B is negative

27.



C is positive and D is negative.


Answer: See diagrams above.

The direction of an electric field is the direction which a positively charged test charge would be accelerated. The electric field lines point in this same direction. As such, electric field lines are always directed away from + charges and towards - charges (or infinity). From this principle, it can be reasoned that A is positive, B is negative, C is positive and D is negative.



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Electric Field Lines

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Part B: Diagramming and Analysis

28. A neutral conducting sphere is charged by induction using a positively-charged balloon. What will be the charge of the following sphere in step e if steps a - d are followed? _________

Explain how the object acquires this charge. Show the type and location of excess charge on the conducting sphere in each step of the diagram (where appropriate).

Answer: See diagram above.

In b., the presence of the positively-charged balloon serves to polarize the sphere. Electrons in the sphere are drawn towards the positively-charged balloon. This leaves a separation of charge in the sphere, with the excess negative charge on the left and the excess positive charge on the right.

In c., the sphere is touched to ground and electrons from the ground are drawn upwards and into the right side of the sphere. It is the presence of the + charge on the right side of the sphere which attracts these electrons to itself. At this point the sphere is charged negatively.

In d., this excess negative charge is shown distributed on the left side of the sphere, still drawn towards the positively-charged balloon.

In e,. the balloon is pulled away and there is a movement of electrons about the sphere so that the excess negative charge is uniformly distributed.



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29. A negatively-charged balloon is held above (without touching) a neutral electroscope. The presence of the nearby balloon causes the needle of the electroscope to deflect.

a. Draw the location and type of excess charges on the "polarized" neutral electroscope.

b. Explain how the balloon has induced the temporary charge upon the electroscope (i.e., describe the direction of electron movement).

 


Answer: See diagram at right.

The negative balloon repels electrons from the plate to the needle and base. Since like charges are repelled, electrons in the top of the electroscope (the plate) are repelled and forced downwards to the bottom of the electroscope (the needle and base). This causes the temporary separation of charge or polarization of the electroscope.


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Polarization || Charging by Induction


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30. Construct electric field lines around the following configuration of charges. Include at least six lines per charge.

a.

b.

c.

d.


Answer: See diagrams above

Electric field lines begin at either infinity or a + charge and are drawn either to infinity or to - charges. Avoid crossing lines. At the surface of charges, the lines should be directed radially inwards (for - charges) or radially outwards (for + charges).



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Polarization || Charging by Induction

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31. Use an unbroken arrow to show the direction of electron movement in the following situation. The arrow should extend from the source of the electrons to the final destination of the electrons.

a. A positively charged balloon is touched to a neutral conducting sphere.



Electrons move from the sphere to the + balloon.

b. A glass rod is rubbed with a piece of wool. The wool has the greater electron affinity.



Electrons move from the glass rod to the wool.

c. A positively-charged balloon is held near a neutral conducting sphere. The sphere is then touched on the opposite side.



Electrons move from the hand to the sphere.

d. A negatively-charged rod is held near a neutral conducting sphere. The sphere is touched on the opposite side.



Electrons move from the sphere to the hand.



Answer: See diagrams above.

In a, c, and d, the movement of electrons is governed by the principles that opposites attract and like-charged objects repel. Electrons (being negative) are attracted to positively charged objects and repelled by negatively charged objects.

In b, the more electron-loving substance (highest electron affinity) gains the electrons from the object with which it is rubbed.



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32. A negatively-charged object creates an electric field which can be measured at various locations in the region about it. An electric field vector is drawn for point X. Use your understanding of the electric field - distance relationship to draw E vectors for points A - E. (The length of the arrow should be indicative of the relative strength of the E vector.)

 

Answer: See diagram above.

The electric field vector points in the direction which a positive test charge would be accelerated. This would result in a vector directed towards a negatively charged object. Thus all E vectors point towards the negatively-charged object. The E vector is inversely dependent upon square of the distance. Thus, the points which are further away have the smallest E vectors.



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Part C: Short Computations

33. The charge of one electron is -1.6 X 10-19 Coulombs. If a neutral object loses 1.5x106 electrons, then what will be its charge?

Answer: +2.4 x 10-13 C

The overall quantity and type of charge on an object is dependent upon the relative number of protons and electrons in the object. To determine the quantity of charge, multiply the charge of one electron by the number of excess or deficient electrons. The type of the charge is dependent upon whether there are more protons or more electrons present in the object. If there are more protons, then the charge is +; otherwise, assign it a - value.



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34. Express your understanding of Coulomb's law by filling in the following table.

 
Q1 (C or µC)
Q2 (C or µC)
d (m)
Felect (N)

a.

1.6 x 10-13 C
1.6 x 10-13 C
0.50 m
9.2 x 10-16 N

b.

1.6 x 10-7 C
1.6 x 10-7 C
0.12 m
0.017 N

c.

4.65 µC
7.28 µC
0.658 m
0.704 N

d.

3.21 µC
0.875 µC
0.553 m
0.0827 N

e.

3.16 x 10-7 C
3.16 x 10-7 C
1.50 m
4.00 x 10-4 N


Answer: See table above.

The electric force (Felect) is computed using Coulomb's law:

Felect = k•Q1•Q2/d2

where Q1 and Q2 represent the charges on the two objects, d represents the separation distance between the object's centers and k = 9x109 N/m2/C2. This equation can be rearranged to solve for any quantity in the equation. The substitutions and algebra are shown below.

a. Felect = (9x109 N/m2/C2)•(1.6 x 10-13 C)•(1.6 x 10-13 C)/(0.5 m)2 = 9.2 x 10-16 N

b. d = SQRT[k•Q1•Q2/Felect] = SQRT[(9x109 N/m2/C2)•(1.6 x 10-7 C)•(1.6 x 10-7 C)/(0.017 N)] = 0.12 N

c. Felect = (9x109 N/m2/C2)•(4.65 x 10-6 C)•(7.28 x 10-6 C)/(0.658 m)2 = 0.704 N

d. Q2 = (d2•Felect )/(k•Q1) = (0.553 m)2•(0.0827 N)/[(9x109 N/m2/C2)•(3.21x10-6 C)] = 8.75-7 C = 0.875 µC

e. Q = SQRT[(d2 •Felect )/(k)] = SQRT[(1.50 m)•(4.00x10-4 N)/(9x109 N/m2/C2)] = 3.16 x 10-7 C



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35. Charged object A (QA) creates an electric field (E). A positively-charged test charge (qB) is used to measure the electric force (F) at various distances (d) from object A. Express your understanding of electric field strength and electric force by filling in the following table. (Note that the units on charge are fictitious units. Since the value of k is not known in these units, you will have to use your qualitative understanding to answer this question.  That is, you will have to think about relationships - the effect that doubling or quadrupling or halving ... the d or the Q has upon the force and the field strength.)

 
QA (bP)
qB (bP)
d (m)
F (mN)
E (mN/bP)

a.

10.0
0.400
1.00
40.0
100.

b.

10.0
0.800
1.00
80.0 100.

c.

10.0
0.400
2.00 10.0
25.0

d.

20.0
0.400
1.00
80.0
200.

e.

20.0
0.400
2.00
20.0 50.0

f.

20.0
0.400
0.500
320. 800.

g.

20.0
1.20
3.00
26.7 22.2


Answer: See table above.

This problem requires you to use the equations as a guide to thinking about how an alteration in one variable would effect the electric force and the electric field. The unique set of units on charge and force mean that you will not be able to plug-and-chug (unless you somehow develop a set of conversion factors between these charge units and force units and the conventional Newtons and Coulombs).

a. The electric field is the force per charge on the test charge. E = F/q = 40 mN/0.40 bP = 100 mN/bP

b. Since electric field does not depend upon the charge of the test charge, the E value at this same distance (compared to row a) will be the same value - 100 mN/bP. With twice the test charge (compared to row a), there will be twice the force - F = 80 N. Now check your thinking to see if E = F/q for your answers; if it does, then you've done well.

c. Here is some complex thinking: the charge creating the field is the same as in row a and row b. But the electric field strength is one-fourth the value; so the distance must have changed. Twice the distance would result in one-fourth the E value. So the distance is 2.0 m. Since electric F also depends inversely upon the square of distance, the electric force must be one-fourth the value of row a (where the qB was the same value).

d. Since the electric field is the force per charge on a test charge, the equation F = E•q can be written. Now substitute values for E and q to solve for F. It is possible to approach this differently without a plug-and-chug approach. Compare row d to row a. The distance is identical but the charge of object A is doubled; doubling the charge would result in a doubling of the force. That is, the new force should be twice the original 40 N value of row a.

e. Compare row e to row d. The only difference between these two rows is the distance in row e is two times row d. Both electric force (F) and electric field (E) are inversely related to the square of the distance. So a doubling of d will result in a fourfold decrease in the value of both F and E. Divide the row d values by 4 to determine the row e values.

f. Once more, compare row f to row d. The only difference between these two rows is the distance in row f is one-half row d. Both electric force (F) and electric field (E) are inversely related to the square of the distance. So a halving of d will result in a fourfold increase in the value of both F and E. Multiply the row d values by 4 to determine the row f values.

g. Now compare row g to row d. In row g, there is triple the test charge and triple the distance. Tripling test charge will triple F and tripling distance will make F one-ninth the original value. So to determine the electric force, take the original (i.e., row d) value of 80 N and multiply by 3 and divide by 9; the result is approximately 26.7 N. To determine E, recognize that alterations in the test charge does not effect the electric field. The electric field in row g will be one-ninth the value of row d - approximately 22.2 N. As a check, divide the force by the test charge to see if indeed it is the same as the electric field value.



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36. The following questions check your qualitative understanding of the variables effecting electric field and electric force. Use your understanding to fill in the blanks.

  1. At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is doubled, then the repulsive force would be 0.0800 N
  2. At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is halved, then the repulsive force would be 1.28 N.
  3. At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is decreased by a factor of 3 and the charge on one of the balloons is doubled, then the repulsive force would be 5.76 N
  4. At a separation distance of 0.500-m, two like-charged balloons experience a repulsive force of 0.320 N. If the distance is is increased by a factor of 3 and the charge on both of the balloons is doubled, then the repulsive force would be 0.142 N.
  5. At a location of 0.200 m from object A, the electric field strength is 0.0500 N/C. If the electric field strength were measured at a location of 0.400 m from object A, then the field strength would be 0.0125 N/C.
  6. At a location of 0.200 m from object A, the electric field strength is 0.0500 N/C. If the electric field strength were measured at a location of 1.00 m from object A, then the field strength would be 0.00200 N/C.
  7. At a location of 0.200 m from object A, the electric field strength is 0.0500 N/C. If the electric field strength were measured using a test charge with twice the charge at a location of 0.200 m from object A, then the field strength would be 0.0500 N/C.
  8. At a location of 0.200 m from object A, the electric field strength is 0.0500 N/C. If the electric field strength were measured using a test charge with twice the charge at a location of 0.100 m from object A, then the field strength would be 0.200 N/C.

 

Answer: See sentences above.

a. If the distance is doubled, then the electric force is one-fourth the original value: 0.320 N/4 = 0.0800 N.

b. If the distance is decreased by a factor of 2 (halved), then the electric force is four times the original value: 0.320 N * 4 = 1.28 N.

c. Decreasing the distance by a factor of 3 will increase the force by a factor of 9; increasing the charge by a factor of 2 will increase the force by a factor of 2. So take the original force adjust accordingly: 0.320 N * 9 * 2 = 5.76 N

d. Increasing the distance by a factor of 3 will decrease the force by a factor of 9; increasing both the charges by a factor of 2 will increase the force by a factor of 2*2. So take the original force adjust accordingly: 0.320 N / 9 * 2 * 2 = 0.142 N

e. If the distance is increased by a factor of 2 (from 0.200 m to 0.400 m) then the electric field strength is decreased by a factor of 4. So take the original E value and adjust accordingly: 0.050 N/C / 4 = 0.0125 N/C.

f. If the distance is increased by a factor of 5 (from 0.200 m to 1.000 m) then the electric field strength is decreased by a factor of 25. So take the original E value and adjust accordingly: 0.0500 N/C / 25 = 0.00200 N/C.

g. Altering the amount of charge on the test charge will not effect the electric field strength at the same location. So the E value remains the same.

h. Altering the amount of charge on the test charge will not effect the electric field strength at the same location. But decreasing the distance by a factor of 2 will increase the electric field strength by a factor of 4. So take the original E value and adjust accordingly: 0.0500 N/C * 4 = 0.200 N/C.



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Part D: Problem-Solving

37. Two objects are located along the y-axis. Object A has a charge of +25 µC and is located at the origin. Object B has a charge of -16 µC and is located a distance of 62 cm above object A. Determine the y-coordinate location where the electric field is zero.

Answer: 310 cm

This problem involves determining a location along the axis where two vectors are directed in opposite direction and have the same magnitude. At such a location the electric field created by object A equals the electric field created by object B.

EA = EB

kQA/RA2 = kQB/RB2

A diagram is always a useful way of approaching the problem. Since A is positively charged, the electric field created by A is directed away from it. Since B is negatively-charged, the electric field created by B is directed towards. Thus, some directional reasoning would lead to the conclusion that the location along the axis must be outside of the two charges. And since object B has the smaller charge and creates the smaller E field, the location will be closer to object B than to object A. That is, what object B lacks in terms of charge it must make up for in terms of distance in order to have an equal field to object A's. The results of this reasoning are shown in the diagram at the right.

The location of 0 electric field is a distance of x cm from the -16 µC charge and a distance of 62+x cm from the +25µC charge. These distances and charge values can be inserted into the equation above and algebra can be used to solve for x. (Note that the + and - sign has been dropped from the equations; directional meaning has been accounted for in the reasoning above.)

k•(25 µC)/(62 + x)2 = k•(16 µC)/(x)2

Divide each side through by k.

(25 µC)/(62 + x)2 = (16 µC)/(x)2

Take the square root of each side of the equation and then cross-multiply.

5/(62 + x) = 4/x

5x = 4•(62 + x) = 248 + 4x

x = 248

Since x represents the distance from object B and since object B is at a y-coordinate of 62 cm, the answer to the problem is 310 cm.



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38. Three objects are located along the x-axis. Object A with a charge of +5.6 µC is located at the origin. Object B has a charge of -4.2 µC and is located at the -1.2 m location. Object C has a charge of +7.7 µC and is located at the +2.4 m location. Determine the magnitude and direction of the net electric force acting upon object A.

Answer: 0.21 N, left

Object A experiences two forces - the force of object B attracting it and pulling it leftward (FBA) and the force of object C repelling it and pushing it leftward (FCA). The diagram below depicts the situation and the forces.

The net electric force upon object A is simply the vector sum of these two forces. The strategy involves computing each force (FBA and FCA ) individually and then adding them together as vectors. The calculations of the two forces are shown below. (Note that the + and - signs have been dropped from the values on the charge since the directional information has already been accounted for in the analysis above.)

FBA = k•QA•QB/d2 = (9.0x109 N/m2/C2)•(5.6x10-6 C)•(4.2 x10-6 C)/(1.2 m)2 = 0.147 N, left

FCA = k•QA•QC/d2 = (9.0x109 N/m2/C2)•(5.6x10-6 C)•(7.7 x10-6 C)/(2.4 m)2 = 0.067 N, left

Now the net electric force on object A can be determined by adding the two forces resulting from the interactions with A and with B. The net force is 0.21 N, left. (0.214 N rounded to two significant digits)



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39. Object A has a x-y coordinate position of (+5.0, 0.0). Object B has a x-y coordinate location of (0.0, +4.0). If object A has a charge of -5.8 µC and object B has a charge of +8.9 µC, then what is the resultant electric field strength at the origin.

Answer: 5.4 x103 N/C, 293 degrees counter-clockwise from east

Both charged object A and charged object B create an electric field, the strength of which is dependent upon the distance from the charged object. This problem involves computing the electric field strength as created by both charges individually and as experienced at the origin (EA and EB). Since electric field strength is a vector, it can be added using rules of vector addition. An electric field vector always has a direction associated with the direction which a + test charge would be forced. Since object A is a negatively charged vector, the vector EA is directed towards it (i.e., rightwards). Since object B is a positively charged vector, the vector EB is directed away from it (i.e., downwards). Once computed individually, EA and EB will be added.

The diagram at the right depicts the situation. The calculations below represent the mathematics.

EA = kQA/RA2 = (9.0x109 N/m2/C2)•(5.8x10-6 C)/(5.0 m)2 = 2.088 x 103 N/C

EB = kQB/RB2 = (9.0x109 N/m2/C2)•(8.9x10-6 C)/(4.0 m)2 = 5.006 x 103 N/C

Now the two E values will be added as vectors. The sketch at the right depicts the addition. Since they are at right angles to each other, the Pythagorean theorem will be used to determine the magnitude and trigonometry will be used to determine the direction.

Enet = SQRT[(2.088 x 103 N/C)2 + (5.006 x 103 N/C)2]

Enet = SQRT[2.942 x 106 N2/C2] = 5.4 x103 N/C

Theta = invtan(EA/EB) = invtan[(2.088 x 103 N/C)/(5.006 x 103 N/C)] = 22.6 deg.

So the direction of the vector is ~23 degrees to the right of down (east of south) or simply 293 degrees counter-clockwise from east.



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40. For the situation described in Question #39, what would be the magnitude of the net electric force upon a +1.3 µC charge if placed at the origin.

Answer: 0.0070 N

This is simply a matter of using the definition of electric field - the electric field is the force per coulomb of charge. Expressed mathematically,

E = F/q

and rearranged, the force is given by the equation

F = E•q

By substitution,

F = (5.4 x103 N/C)•(1.3x10-6 C) = 0.0070 N


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41. Alpha particles (i.e., Helium nucleus) have a molar mass of 4.002 g/mol and consist of two protons and two neutrons.

  1. Determine the charge of one alpha particle in units of Coulombs and the mass of one alpha particle in units of kg.
  2. Suppose that Ernie Rufferthord (not to be confused with the scientist of gold foil fame) wishes to suspend an alpha particle in midair by attracting it to a bundle of electrons held a distance of 1.00-m above the alpha particle. How many electrons would Ernie need in his bundle to accomplish such an amazing feat? 

Answer: approx. 141 electrons

a. This involves a conversion using Avogadro's number and the molar mass:

mass = 1 alpha particle • (1 mol/6.02 x 1023 alpha particles) • (4.002 g/1 mol) • (1 kg/1000 g)

mass = 6.65 x 10-27 kg

 

b. Ernie wishes to use the force of electrical attraction between the alpha particle and the electron bundle to balance the downward pull of gravity upon the alpha particle. So

Felect = Fgrav

k • Qalpha particle • Qe- bundle/R2 = malpha particle • g

The charge on the alpha particle can be determined from its composition - it contains 2 protons (and 2 neutral neutrons), each having a charge of 1.6x10-19 C. So the charge of an alpha particle is 3.2x10-19 C. The only unknown remaining is the charge on the electron bundle. Rearrangement of the equation yields the following expression for the charge on the electron bundle:

Qe- bundle = malpha particle • g • R2 / (k • Qalpha particle)

Substitution of known values into this equation yields 2.26x10-17 C as the charge on the electron bundle. With each electron contributed 1.6x10-19 C per electron, there must be approximately 141 electrons in the bundle.



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42. A 1.19-gram charged balloon hangs from a 1.99-m string which is attached to the ceiling. A Van de Graaff generator is located directly below the location where the string attaches to the ceiling and is at the same height as the balloon. The string is deflected at an angle of 32.0 degrees from the vertical due to the presence of the electric field. Determine the charge on the Van de Graaff generator if the charge on the balloon is 2.27x10-12 C.

Answer: 0.397 C

The balloon in this problem is experiencing an electric force resulting from the interaction with the Van de Graaff generator. Diagram A illustrates the situation. The amount of charge on the Van de Graaff generator will be related to the distance of separation (R), the electrical force of repulsion, and the charge on the balloon (which is given). So the problem-solving strategy involves determining the electrical force of repulsion using a force analysis and then using Coulomb's law to determine the charge on the Van de Graaff generator.

Diagram B above shows a free-body diagram for the balloon. The three forces must balance since the balloon is at rest. Therefore, the x-component of the tension force (Fx in Diagram C) must equal the electrical force (Felect). And similarly, the y-component of the tension force (Fy in Diagram C) must equal the gravity force (Fgrav). A trigonometric function relating the angle of 32.0 degrees to these components of forces can be written.

tan(32.0 deg) = opposite/adjacent = Fx /Fy

And due to the equivalency between the components and the other two forces, it can also be stated that:

tan(32.0 deg) = Felect / Fgrav

The equation can be rearranged to solve for Felect; the expression for Fgrav and the related values can be substituted.

Felect = Fgrav • tan(32.0 deg) = m•g•tan(32.0 deg) = (0.00119 kg)•(9.8 m/s/s)•tan(32.0 deg)

Felect = 7.29 x 10-3 N

Now the separation distance (R) between the balloon and the generator can be determined. Observe in Diagram A that the distance R is related to the length of the string and the angle with the vertical. Using the sine function, R can be determined as follows:

sin(32.0 deg) = opposite/hypotenuse = R/1.99 m

R = 1.99 m • sin(32.0 deg) = 1.05 m

Now with Felect, R, and Q1 known, Q2 (the charge on the Van de Graaff generator) can be determined.

Felect = k•Q1•Q2/R2

By rearranging, the equation becomes

Q2 = Felect • R2 / (k•Q1) = (7.29 x 10-3 N) • (1.05)2 / [ (9.0x109 N•m2/C2) • (2.27x10-12 C)]

Q2 = 0.397 C



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43. Ignoring trace elements, a typical elemental composition (by mass) of the human body is as follows:

10.5 % H
66.3 % O
19.9 % C
3.3 % N
  1. Use these percentages, the molar mass values, Avogadro's number, and the atomic numbers to determine the total number of electrons (and protons) in a 73-kg human body (160 pounds).
  2. If these electrons and protons were placed 100 m apart (the distance of approximately one football field), then what would be the force of electrical attraction between them. 

Answer: See answers below.

a. How many electrons are in your body? It's kind of a neat question; and remarkably, a student of Chemistry and Physics knows enough to come up with an estimate.

The strategy involves first using the mass of the body, the percent mass values, the atomic mass values, and Avogadro's number to first determine the number of atoms of each element in the body; then use the atomic number of each element to determine the number of electrons contributed by each element (assuming each atom is neutral); finally add all the electrons to determine the total number of electrons. Here it goes:

When added, there are 24.3x1027 electrons (and a presumed equal number of protons).

 

b. The force of attraction of these protons and electrons at a separation distance of 100 m can easily be calculated if the net + and net - charge is known. Simply multiply the number of electrons by the charge of one electron (1.6x10-19 C); do similarly for the protons. The charge ends up being 3.9x109 C. Now Coulomb's law can be used to compute the electric force of attraction:

Felect = k•Q1•Q2/R2 = (9.0x109 N•m2/C2)•(3.9x109 C)•(3.9x109 C)/(100 m)2 = 1.4x1025 N


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