Conservation of Electric Charge - College Physics

1. A conducting sphere with radius $R$ and total charge $+Q$ is surrounded by a concentric conducting shell with inner radius $2R$ and outer radius $3R$. The shell is initially uncharged. When electrostatic equilibrium is established, what is the charge on the inner surface of the shell?
   Answer: $-Q$ By Gauss's law and the properties of conductors, the electric field inside a conductor must be zero. This requires that the total charge enclosed by any Gaussian surface within the conducting shell be zero. Since the inner sphere has charge +Q, the inner surface of the shell must have charge -Q to make the total enclosed charge zero. The outer surface of the shell will have charge +Q to conserve the total charge of the shell (which was initially zero).
2. A physics student performs an experiment where a negatively charged rod is brought near (but not touching) a neutral electroscope, causing the leaves to diverge. When the rod is removed, the leaves collapse. What happened to the total charge of the electroscope during this process?
   Answer: The total charge remained zero throughout the entire process This question tests your understanding of electrostatic induction, a fundamental concept where charges redistribute within a conductor without any actual charge transfer to or from the object. When you bring a negatively charged rod near the neutral electroscope, the rod's electric field causes the free electrons in the metal electroscope to redistribute. The excess electrons on the rod repel electrons in the electroscope, pushing them away from the rod's location. This creates a charge separation: the side of the electroscope nearest the rod becomes positively charged (electron deficit), while the far side becomes negatively charged (electron excess). The leaves diverge because they both acquire the same type of charge and repel each other. Crucially, no electrons actually leave or enter the electroscope—they just rearrange within it. Answer A is correct because charge is conserved throughout this process. The electroscope started neutral and remains neutral overall, even though charges are temporarily separated. Answer B is wrong because the electroscope never gains extra negative charge—its electrons just redistribute internally. Answer C is incorrect because while electron repulsion does occur, this doesn't create a net positive charge on the electroscope, only a temporary separation. Answer D is wrong because the charge doesn't oscillate; it simply separates when the rod approaches and redistributes evenly when removed. Remember: in electrostatic induction problems, always distinguish between charge separation (redistribution) and actual charge transfer. The total charge of an isolated conductor remains constant during induction.
3. A technician working with electronic components accidentally touches a charged capacitor plate with a grounded probe. Before contact, the plate had $+8.0\times 10^{-6}$ C of charge. After the probe contact, the plate has zero charge. What happened to the original charge on the plate?
   Answer: The charge flowed through the probe to ground, conserving total charge in the larger system This question tests your understanding of charge conservation, one of the fundamental principles in electrostatics. When you encounter problems involving charge transfer or grounding, always think about where the charge goes—it can't simply disappear. When the grounded probe touches the charged capacitor plate, the excess positive charge flows through the probe to ground. Ground acts as an infinite reservoir that can accept or supply charge without changing its own potential. The $+8.0\times 10^{-6}$ C doesn't vanish; it spreads out into the Earth's vast conducting mass where it becomes negligible. This process conserves the total charge in the system—the charge just redistributes from a small, concentrated location to an enormously large one. Looking at the wrong answers: Option B incorrectly suggests charge can be destroyed when converted to heat. While some energy is dissipated as heat during discharge, charge itself is conserved—energy and charge are different quantities. Option C misunderstands what grounding does; the charge doesn't hide on the plate, it actually leaves the plate entirely. Option D describes charge neutralization, but this would require negative charges to combine with positive ones on the plate itself, which isn't what happens during grounding. Remember this key distinction: grounding removes charge by providing a path for it to flow away, while neutralization involves opposite charges combining in the same location. When you see grounding problems, always trace where the charge flows rather than assuming it disappears.
4. A student uses a plastic comb to pick up small pieces of paper after running the comb through dry hair. If the comb acquires $-2.4 \times 10^{-9}$ C of charge during this process, what was the net change in the number of electrons on the comb?
   Answer: An increase of $1.5 \times 10^{10}$ electrons When you encounter electrostatics problems involving charge and electrons, remember that electric charge is quantized—it comes in discrete packets equal to the elementary charge of a single electron or proton. To find the change in electron number, you need to divide the total charge by the elementary charge of one electron. The elementary charge is $e = 1.6 \times 10^{-19}$ C. Since the comb acquired $-2.4 \times 10^{-9}$ C, the calculation is: $\text{Number of electrons} = \frac{|-2.4 \times 10^{-9} \text{ C}|}{1.6 \times 10^{-19} \text{ C/electron}} = 1.5 \times 10^{10} \text{ electrons}$ The negative charge means the comb gained electrons (electrons have negative charge), so this represents an increase of $1.5 \times 10^{10}$ electrons, making A correct. Looking at the wrong answers: B suggests a decrease in electrons, but negative charge means electron gain, not loss. C shows $2.4 \times 10^{-9}$ electrons, which incorrectly uses the charge value as the number of electrons—this ignores that you must divide by the elementary charge. D claims no change, which contradicts the given charge acquisition. Study tip: Always remember the elementary charge constant ($1.6 \times 10^{-19}$ C) and that negative charge means excess electrons while positive charge means electron deficiency. The relationship $Q = ne$ (where $n$ is the number of electrons) is fundamental to electrostatics problems.
5. In a thundercloud, charge separation occurs with the bottom of the cloud becoming negatively charged and the top becoming positively charged. If the bottom region acquires a charge of $-40 \text{ C}$ and we assume this charge came entirely from electrons moved from the top region, approximately how many electrons were transferred from the top to the bottom of the cloud?
   Answer: $2.5 \times 10^{20}$ The number of electrons transferred equals the magnitude of charge divided by the elementary charge: $N = \frac{|Q|}{e} = \frac{40 \text{ C}}{1.6 \times 10^{-19} \text{ C}} = 2.5 \times 10^{20}$. Each electron carries charge $-1.6 \times 10^{-19} \text{ C}$, so this many electrons gives the total charge of -40 C in the bottom region. By conservation of charge, the top region loses this many electrons and becomes +40 C.
6. Two identical metal spheres are suspended by insulating threads. Sphere X has a charge of $+12 \mu\text{C}$ and sphere Y has a charge of $-4 \mu\text{C}$. The spheres are brought into contact and then separated. Which statement correctly describes the redistribution of charge?
   Answer: Each sphere will have a charge of $+4 \mu\text{C}$ after contact When two conducting spheres touch, you're dealing with charge redistribution based on conservation of charge and the principle that conductors in contact reach the same electric potential. Since these are identical metal spheres, when they touch, the total charge redistributes equally between them. Start by finding the total charge: $(+12 \mu\text{C}) + (-4 \mu\text{C}) = +8 \mu\text{C}$. This total charge must be conserved - it cannot be created or destroyed. When identical conductors are brought into contact, they share this total charge equally, so each sphere gets $\frac{+8 \mu\text{C}}{2} = +4 \mu\text{C}$. Choice A is correct because it properly applies charge conservation and equal distribution for identical conductors. Choice B incorrectly suggests an unequal distribution ($+8 \mu\text{C}$ and $0 \mu\text{C}$). While this does conserve total charge, identical spheres in contact must have equal charge distributions - there's no physical reason for one to retain more charge than the other. Choice C falls into the neutralization trap, assuming positive and negative charges simply cancel out completely. However, sphere X has more positive charge than sphere Y has negative charge, so complete neutralization is impossible. Choice D shows an unequal split ($+6 \mu\text{C}$ and $+2 \mu\text{C}$) that conserves total charge but violates the equal distribution principle for identical conductors. Remember: when identical conductors touch, always find the total charge first, then divide equally. The key word "identical" tells you the final charges must be equal.
7. A plastic rod is rubbed with a wool cloth, causing the rod to acquire a charge of $-4.8 \times 10^{-8} \text{ C}$. During the rubbing process, some electrons are transferred between the materials. If each transferred electron contributes its full elementary charge to the imbalance, how many electrons were transferred, and in which direction?
   Answer: $3.0 \times 10^{11}$ electrons transferred from the wool to the rod When you encounter charge transfer problems, remember that charge quantization means all charge comes in discrete packets equal to the elementary charge of an electron: $e = 1.6 \times 10^{-19} \text{ C}$. To find how many electrons were transferred, divide the total charge by the elementary charge: $n = \frac{|q|}{e} = \frac{4.8 \times 10^{-8}}{1.6 \times 10^{-19}} = 3.0 \times 10^{11}$ electrons. Since the rod acquired a negative charge, it must have gained electrons (negative charges). In triboelectric charging, electrons always move from one material to another - they transferred from the wool to the rod. Looking at the wrong answers: Choice A gives the correct direction but calculates $n = \frac{4.8 \times 10^{-8}}{3.2 \times 10^{-19}}$, suggesting someone doubled the elementary charge. Choice B has the correct number but wrong direction - if electrons moved from rod to wool, the rod would become positive, not negative. Choice C makes a critical error by dividing the elementary charge by the total charge instead of vice versa: $\frac{1.6 \times 10^{-19}}{4.8 \times 10^{-8}} = 3.3 \times 10^{-12}$, though even this doesn't match their answer exactly. The correct answer is D: $3.0 \times 10^{11}$ electrons transferred from wool to rod. Study tip: Always check that your direction makes sense with the final charge sign. Gaining electrons = negative charge, losing electrons = positive charge. This sign check catches many mistakes on electrostatics problems.
8. An electroscope consists of a metal knob connected to two thin metal leaves that can move apart when charged. Initially, the electroscope is uncharged and the leaves hang straight down. A positively charged rod is brought near (but not touching) the knob. While the rod is nearby, the knob is momentarily touched by a finger and then the finger is removed while the rod remains in position. Finally, the rod is moved away. Describe the final state of the electroscope.
   Answer: The electroscope becomes negatively charged and the leaves spread apart This is charging by induction. When the positive rod approaches, it attracts electrons to the knob, leaving the leaves positive. Touching the knob allows electrons to flow from ground through the finger to neutralize the positive charge on the leaves. When the finger is removed while the rod is still present, the electroscope retains excess electrons. When the rod is removed, these electrons redistribute throughout the electroscope, including the leaves. The leaves both become negatively charged and repel each other, causing them to spread apart.
9. Two identical conducting spheres are connected by a thin conducting wire. Sphere 1 initially has charge $+Q$ and sphere 2 initially has charge $-3Q$. After equilibrium is established through the wire, what is the charge on sphere 1?
   Answer: $-Q$ When two conducting objects are connected by a conducting wire, charge flows until both objects reach the same electric potential. For identical conducting spheres, this means charge distributes equally between them. Start by applying conservation of charge. The total charge in the system must remain constant: initial charge = $+Q + (-3Q) = -2Q$. After equilibrium, this $-2Q$ total charge splits equally between the two identical spheres, so each sphere gets $\frac{-2Q}{2} = -Q$. Therefore, sphere 1 ends up with charge $-Q$, making choice (A) correct. Let's examine why the other answers are wrong. Choice (B) suggests sphere 1 keeps its original charge of $+Q$, which ignores that charge must redistribute when conductors are connected. Choice (C) gives $-2Q$, which would mean all the charge ended up on sphere 1 while sphere 2 has zero charge - this violates the equal potential requirement for identical connected conductors. Choice (D) suggests zero charge on sphere 1, which would mean sphere 2 has all $-2Q$ charge, again violating equal charge distribution. The key insight is that identical connected conductors always share charge equally because they must have the same potential. When you see problems involving connected conducting objects, immediately check if they're identical. If so, calculate the total charge, then divide equally among the objects. This principle applies whether the objects start with the same or different charges.
10. Four conducting spheres A, B, C, and D are initially isolated. Their charges are: A has $+12 \text{ nC}$, B has $-6 \text{ nC}$, C has $+3 \text{ nC}$, and D has $-9 \text{ nC}$. First, A and B are touched together and separated. Then, C and D are touched together and separated. Finally, the sphere that originally was A is touched to the sphere that originally was C, and they are separated. What is the final charge on the sphere that was originally sphere A?
    Answer: $0 \text{ nC}$ When conductors touch, charge redistributes equally between them until they reach the same electric potential. For identical conducting spheres, this means the total charge splits evenly between the two spheres. Let's trace through each step systematically. Initially: A has $+12 \text{ nC}$, B has $-6 \text{ nC}$, C has $+3 \text{ nC}$, and D has $-9 \text{ nC}$. Step 1: A and B touch. Total charge = $+12 + (-6) = +6 \text{ nC}$. This splits evenly, so each gets $+3 \text{ nC}$. Now A has $+3 \text{ nC}$ and B has $+3 \text{ nC}$. Step 2: C and D touch. Total charge = $+3 + (-9) = -6 \text{ nC}$. This splits evenly, so each gets $-3 \text{ nC}$. Now C has $-3 \text{ nC}$ and D has $-3 \text{ nC}$. Step 3: A and C touch. Total charge = $+3 + (-3) = 0 \text{ nC}$. This splits evenly, so each gets $0 \text{ nC}$. Therefore, A ends with $0 \text{ nC}$. The answer is **(D) $0 \text{ nC}$**. **(A) $-1.5 \text{ nC}$ and (B) $+1.5 \text{ nC}$ likely come from calculation errors, perhaps incorrectly handling the charge distribution in one of the steps. (C) $+3 \text{ nC}$** represents A's charge after the first contact but ignores the final step. Strategy tip: Always work through charge redistribution problems step-by-step, carefully tracking each sphere's charge after every contact. The key principle is that identical conductors always split charge equally.
11. In a laboratory demonstration, two conducting spheres are mounted on insulating stands. Sphere A has charge $+12\mu\text{C}$ and sphere B has charge $+3\mu\text{C}$. A conducting wire is briefly connected between the spheres and then removed. If sphere A has twice the radius of sphere B, what is the final charge on sphere A?
    Answer: $+10\mu\text{C}$ When conducting spheres are connected by a wire, charge redistributes until both spheres reach the same electric potential. This is a key principle in electrostatics—charge flows from higher to lower potential until equilibrium is reached. For conducting spheres, the potential is $V = \frac{kQ}{R}$, where $Q$ is the charge and $R$ is the radius. At equilibrium, $V_A = V_B$, so: $\frac{kQ_A}{R_A} = \frac{kQ_B}{R_B}$ This simplifies to $\frac{Q_A}{R_A} = \frac{Q_B}{R_B}$ Since sphere A has twice the radius of sphere B ($R_A = 2R_B$), we get: $\frac{Q_A}{2R_B} = \frac{Q_B}{R_B}$ Therefore: $Q_A = 2Q_B$ The total charge is conserved: $Q_A + Q_B = 12\mu\text{C} + 3\mu\text{C} = 15\mu\text{C}$ Substituting $Q_A = 2Q_B$: $2Q_B + Q_B = 15\mu\text{C}$ $Q_B = 5\mu\text{C}$ and $Q_A = 10\mu\text{C}$ Answer A ($+10\mu\text{C}$) is correct. Answer B ($+7.5\mu\text{C}$) incorrectly assumes charge splits proportionally by mass or volume. Answer C ($+6\mu\text{C}$) wrongly assumes charge distributes inversely with radius. Answer D ($+9\mu\text{C}$) might result from calculation errors or misapplying the radius relationship. Study tip: Remember that larger conducting spheres hold more charge at the same potential. The charge ratio equals the radius ratio, not the inverse.
12. Three small conducting spheres A, B, and C are initially isolated from each other. Sphere A has charge $+Q$, sphere B has charge $-2Q$, and sphere C is uncharged. First, spheres A and B are brought into contact and then separated. Next, sphere A is brought into contact with sphere C and then separated. What is the final charge on sphere C?
    Answer: $-\frac{Q}{4}$ When conducting spheres touch, total charge distributes equally. Initially: A has +Q, B has -2Q, C has 0. First contact (A and B): Total charge = (+Q) + (-2Q) = -Q. After separation, each has -Q/2. So A has -Q/2, B has -Q/2, C still has 0. Second contact (A and C): A has -Q/2, C has 0. Total = -Q/2. After separation, each has (-Q/2)/2 = -Q/4. Therefore, sphere C has final charge -Q/4.
13. In a photocopying machine, a selenium drum initially has no charge. During operation, it acquires a charge of $+4.8 \times 10^{-9}$ C through a charging process. How many electrons were removed from the drum during this charging?
    Answer: $3.0 \times 10^{10}$ When you encounter charge and electron problems, remember that charge is quantized—it comes in discrete packets equal to the elementary charge of a single electron or proton. To find how many electrons were removed, you need to divide the total positive charge by the charge of a single electron. The fundamental charge is $e = 1.6 \times 10^{-19}$ C. Since the drum gained a positive charge of $+4.8 \times 10^{-9}$ C, this means electrons (which carry negative charge) were removed. Number of electrons = $\frac{\text{Total charge}}{\text{Charge per electron}} = \frac{4.8 \times 10^{-9}}{1.6 \times 10^{-19}} = 3.0 \times 10^{10}$ Choice A ($3.0 \times 10^{10}$) is correct—this represents the actual number of electrons removed. Choice B ($4.8 \times 10^{-9}$) is simply the total charge given in the problem, not the number of electrons. This represents a failure to perform the necessary division. Choice C ($7.68 \times 10^{-28}$) results from incorrectly multiplying the total charge by the elementary charge instead of dividing: $4.8 \times 10^{-9} \times 1.6 \times 10^{-19}$. Choice D ($1.6 \times 10^{-19}$) is just the elementary charge constant itself, showing confusion about what the question is asking. Study tip: Always remember that finding the number of charge carriers requires dividing total charge by elementary charge. The key relationship is: $N = \frac{Q}{e}$ where N is the number of electrons, Q is total charge, and e is elementary charge.
14. In a physics demonstration, a Van de Graaff generator transfers electrons from its belt to ground, leaving the dome with a positive charge. If the generator removes $3.2 \times 10^{12}$ electrons from the dome in 4.0 seconds, what is the average current flowing from the dome to ground during this time?
    Answer: $1.3 \times 10^{-7} \text{ A}$ Current is the rate of charge flow: $I = \frac{Q}{t}$. The total charge transferred is the number of electrons times the elementary charge: $Q = (3.2 \times 10^{12})(1.6 \times 10^{-19} \text{ C}) = 5.12 \times 10^{-7} \text{ C}$. The average current is $I = \frac{5.12 \times 10^{-7} \text{ C}}{4.0 \text{ s}} = 1.28 \times 10^{-7} \text{ A} \approx 1.3 \times 10^{-7} \text{ A}$.
15. A student rubs a glass rod with silk, transferring $1.8 \times 10^{12}$ electrons from the glass to the silk. If the glass rod initially had no net charge, what is the magnitude of charge on the silk after rubbing?
    Answer: $2.88 \times 10^{-7}$ C When you encounter electrostatics problems involving charge transfer, remember that charge is quantized—it comes in discrete packets equal to the elementary charge of a single electron or proton. Since electrons are transferred from glass to silk, the silk gains negative charge. To find the total charge, you multiply the number of electrons by the elementary charge: $q = n \times e$, where $n = 1.8 \times 10^{12}$ electrons and $e = 1.6 \times 10^{-19}$ C per electron. $q = (1.8 \times 10^{12}) \times (1.6 \times 10^{-19} \text{ C}) = 2.88 \times 10^{-7} \text{ C}$ This confirms answer A is correct. Answer B ($1.8 \times 10^{12}$ C) represents the trap of using the number of electrons as the charge magnitude—forgetting that each electron carries only $1.6 \times 10^{-19}$ C, not 1 C. This would be an enormous charge in everyday terms. Answer C ($1.6 \times 10^{-19}$ C) gives you the charge of just one electron, suggesting you forgot to multiply by the total number of electrons transferred. Answer D ($2.88 \times 10^{7}$ C) results from an incorrect exponent calculation—likely adding exponents incorrectly when multiplying the scientific notation (getting $10^{-7}$ wrong as $10^{7}$). Study tip: Always remember the elementary charge constant $e = 1.6 \times 10^{-19}$ C, and when dealing with multiple particles, multiply by the number of particles. Double-check your exponent arithmetic in scientific notation calculations.
16. A Van de Graaff generator transfers electrons from its base to ground, leaving the spherical dome with a positive charge. If $3.2 \times 10^{14}$ electrons are removed from the dome, and each electron carries a charge of magnitude $1.6 \times 10^{-19}$ C, what is the net charge acquired by the dome?
    Answer: $+5.12 \times 10^{-5}$ C When you encounter electrostatic charge problems, remember that charge is quantized—it comes in discrete packets equal to the elementary charge of an electron or proton. The key is calculating total charge by multiplying the number of charge carriers by the charge per carrier. Since the Van de Graaff generator removes electrons from the dome, you're left with a deficit of negative charge, making the dome positively charged. To find the total charge, multiply the number of electrons removed by the charge magnitude per electron: Total charge = (number of electrons) × (charge per electron) Total charge = $(3.2 \times 10^{14}) \times (1.6 \times 10^{-19} \text{ C})$ Total charge = $5.12 \times 10^{-5} \text{ C}$ Since electrons were removed, the dome becomes positively charged: $+5.12 \times 10^{-5}$ C. Choice A is correct—it properly calculates the magnitude and recognizes the positive sign from electron removal. Choice B gives the correct magnitude but wrong sign; this would be the charge if electrons were added to the dome. Choice C mistakenly uses the number of electrons as the charge value, ignoring that you must multiply by the elementary charge. Choice D results from incorrectly dividing instead of multiplying the given values. Study tip: Always check units and signs in charge problems. When electrons are removed, the object becomes positive; when added, it becomes negative. The elementary charge $1.6 \times 10^{-19}$ C is a fundamental constant you should memorize for physics problems.
17. In an electrostatic painting process, negatively charged paint droplets are sprayed toward a grounded metal car frame. As the droplets approach and stick to the frame, what happens to the total charge of the paint-frame system?
    Answer: The total charge remains constant as electrons redistribute between components When you encounter electrostatic problems, always think about conservation of charge - one of the fundamental laws of physics. Charge cannot be created or destroyed in an isolated system; it can only be transferred from one object to another. In this electrostatic painting scenario, you have an isolated system consisting of negatively charged paint droplets and a grounded metal frame. Initially, the paint carries negative charge while the frame is neutral (grounded simply means it can exchange charge with the earth to maintain zero potential, but once painting begins, treat the paint-frame system as isolated). When the droplets stick to the frame, the electrons don't disappear - they redistribute within the combined system. Answer A correctly identifies that total charge remains constant through electron redistribution. The negative charges that were on the droplets are now part of the paint-frame system, but the algebraic sum of all charges stays the same. Answer B incorrectly suggests charge destruction. "Neutralization" might seem intuitive, but it violates charge conservation. The electrons don't vanish; they're still present in the system. Answer C confuses kinetic energy with charge creation. While the droplets have kinetic energy as they move, energy and charge are completely different quantities. Motion cannot generate charge. Answer D assumes complete neutralization occurs, which would require an external source of positive charge to cancel the negative charges - but no such source exists in this system. Remember: In electrostatics problems, always account for charge conservation first. Charges redistribute, but the total never changes unless charge enters or leaves the system.
18. A neutral metal sphere is brought into contact with a negatively charged rod. After contact, the sphere is found to have a charge of $-8.0 \times 10^{-6}$ C. If the rod originally had a charge of $-12.0 \times 10^{-6}$ C, what is the final charge on the rod?
    Answer: $-4.0 \times 10^{-6}$ C When you encounter problems involving charge transfer between conducting objects, remember that charge is conserved—the total amount of charge before and after contact must remain the same. Initially, you have a neutral metal sphere (charge = 0) and a negatively charged rod with $-12.0 \times 10^{-6}$ C. The total initial charge is $0 + (-12.0 \times 10^{-6}) = -12.0 \times 10^{-6}$ C. After contact, the sphere has $-8.0 \times 10^{-6}$ C. Since charge must be conserved, the rod's final charge equals the total initial charge minus the sphere's final charge: $-12.0 \times 10^{-6} - (-8.0 \times 10^{-6}) = -4.0 \times 10^{-6}$ C. Looking at the wrong answers: Choice B ($-20.0 \times 10^{-6}$ C) incorrectly adds the charges, violating conservation of charge. Choice C ($+4.0 \times 10^{-6}$ C) has the right magnitude but wrong sign—this would require the rod to gain positive charge, which is impossible when both objects start with zero or negative charge. Choice D ($+8.0 \times 10^{-6}$ C) makes the same sign error and uses the sphere's final charge value. The correct answer is A: $-4.0 \times 10^{-6}$ C. Study tip: For any charge transfer problem, write down the conservation equation first: $Q_{initial,total} = Q_{final,total}$. This prevents sign errors and helps you set up the problem correctly. Always check that your final answer makes physical sense—charges don't disappear or change sign without reason.
19. A metal rod is initially electrically neutral. One end of the rod is brought near (but not touching) a positively charged sphere. While the sphere remains in this position, the far end of the rod is momentarily grounded by touching it with a finger, then the finger is removed while the sphere is still nearby. Finally, the charged sphere is moved far away. What is the final state of the rod?
    Answer: The rod becomes negatively charged with the charge distributed uniformly throughout the rod This describes charging by induction. When the positive sphere approaches, electrons in the neutral rod are attracted to the near end, leaving the far end positive. Grounding the far end allows positive charge to flow away (or equivalently, electrons flow from ground to neutralize the positive end). When the finger is removed, the rod retains excess electrons. When the sphere is removed, these electrons redistribute uniformly throughout the conductor, leaving the rod negatively charged overall with uniform distribution.
20. A balloon is rubbed with wool, causing the balloon to become negatively charged with $-3.2 \times 10^{-8}$ C. Assuming the wool was initially neutral, how many excess electrons are now on the balloon?
    Answer: $2.0 \times 10^{11}$ This question tests your understanding of electric charge quantization - the fundamental principle that all electric charge comes in discrete units equal to the elementary charge of a single electron or proton. To find the number of excess electrons, you need to divide the total charge by the charge of a single electron. The elementary charge is $e = 1.6 \times 10^{-19}$ C. Since the balloon has a negative charge of $-3.2 \times 10^{-8}$ C, you calculate: Number of electrons = $\frac{|{-3.2 \times 10^{-8}}|}{1.6 \times 10^{-19}} = \frac{3.2 \times 10^{-8}}{1.6 \times 10^{-19}} = 2.0 \times 10^{11}$ This confirms that answer A is correct. Looking at the wrong answers: B ($3.2 \times 10^{-8}$) is simply the magnitude of the total charge in coulombs - a common trap for students who forget to convert from charge to number of particles. C ($5.12 \times 10^{-27}$) appears to be the result of incorrectly multiplying the charge by the elementary charge instead of dividing, giving you an impossibly small fraction of an electron. D ($1.6 \times 10^{-19}$) is the elementary charge itself - another trap for students who might confuse the fundamental constant with the answer. Study tip: Memorize that the elementary charge is $1.6 \times 10^{-19}$ C. Whenever you see a charge quantization problem, always divide the total charge by this value to find the number of elementary charges involved.

Start this quiz