College Physics Quiz: Conservation Of Electric Charge
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Conservation Of Electric ChargeQuestion 1 of 20

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 C-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?

6.4×10−216.4 \times 10^{-21}
2.5×10202.5 \times 10^{20}
4.0×10194.0 \times 10^{19}
1.6×10211.6 \times 10^{21}
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College Physics Quiz

College Physics Quiz: Conservation Of Electric Charge

Practice Conservation Of Electric Charge in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Conservation Of Electric Charge, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

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 C-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?

  1. 6.4×10−216.4 \times 10^{-21}
  2. 2.5×10202.5 \times 10^{20} (correct answer)
  3. 4.0×10194.0 \times 10^{19}
  4. 1.6×10211.6 \times 10^{21}
Explanation: The number of electrons transferred equals the magnitude of charge divided by the elementary charge: N=∣Q∣e=40 C1.6×10−19 C=2.5×1020N = \frac{|Q|}{e} = \frac{40 \text{ C}}{1.6 \times 10^{-19} \text{ C}} = 2.5 \times 10^{20}. Each electron carries charge −1.6×10−19 C-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.

Question 2

A neutral metal sphere is brought into contact with a negatively charged rod. After contact, the sphere is found to have a charge of −8.0×10−6-8.0 \times 10^{-6} C. If the rod originally had a charge of −12.0×10−6-12.0 \times 10^{-6} C, what is the final charge on the rod?

  1. −4.0×10−6-4.0 \times 10^{-6} C (correct answer)
  2. −20.0×10−6-20.0 \times 10^{-6} C
  3. +4.0×10−6+4.0 \times 10^{-6} C
  4. +8.0×10−6+8.0 \times 10^{-6} C
  5. 00 C
Explanation: 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 −12.0×10−6-12.0 \times 10^{-6} C. The total initial charge is 0+(−12.0×10−6)=−12.0×10−60 + (-12.0 \times 10^{-6}) = -12.0 \times 10^{-6} C. After contact, the sphere has −8.0×10−6-8.0 \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: −12.0×10−6−(−8.0×10−6)=−4.0×10−6-12.0 \times 10^{-6} - (-8.0 \times 10^{-6}) = -4.0 \times 10^{-6} C. Looking at the wrong answers: Choice B (−20.0×10−6-20.0 \times 10^{-6} C) incorrectly adds the charges, violating conservation of charge. Choice C (+4.0×10−6+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 (+8.0×10−6+8.0 \times 10^{-6} C) makes the same sign error and uses the sphere's final charge value. The correct answer is A: −4.0×10−6-4.0 \times 10^{-6} C. Study tip: For any charge transfer problem, write down the conservation equation first: Qinitial,total=Qfinal,totalQ_{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.

Question 3

In a thundercloud, charge separation occurs with the bottom of the cloud becoming negatively charged. If the cloud loses 2.5×10202.5 \times 10^{20} electrons to the ground during a lightning strike, what happens to the charge of the cloud-ground system?

  1. The total charge of the system remains constant throughout the process (correct answer)
  2. The total charge of the system decreases by the amount transferred
  3. The total charge of the system increases due to the high energy involved
  4. The total charge of the system becomes zero after the lightning strike
  5. The total charge of the system doubles because both cloud and ground change
Explanation: When you encounter questions about charge transfer during lightning, think about one of the most fundamental principles in physics: conservation of charge. This principle states that electric charge can neither be created nor destroyed, only transferred from one location to another. During the lightning strike described, 2.5×10202.5 \times 10^{20} electrons move from the negatively charged bottom of the cloud to the ground. This transfer doesn't eliminate these electrons or create new charge - it simply redistributes the existing charge within the cloud-ground system. Before the strike, the cloud had excess electrons (negative charge) and the ground had a corresponding positive charge. After the strike, both the cloud and ground move closer to electrical neutrality, but the total amount of charge in the system remains exactly the same. Answer A is correct because conservation of charge ensures the total charge stays constant throughout the process. Answer B incorrectly suggests that charge is destroyed when electrons transfer - but transfer isn't destruction. Answer C reflects a common misconception that high energy processes can create charge, but energy and charge are entirely different quantities that follow different conservation laws. Answer D wrongly assumes that the system reaches perfect neutrality after one lightning strike, which rarely happens since complete charge balance typically requires multiple discharge events. Remember this key distinction: energy can be converted between forms, but charge can only be moved around. When you see any problem involving charge transfer - whether it's lightning, static electricity, or circuits - always start by applying conservation of charge.

Question 4

A positively charged rod is used to charge a metal sphere by induction. During this process, electrons in the sphere move toward the rod, leaving the far side of the sphere positively charged. If the sphere is then grounded while the rod is still present, what is the final charge on the sphere after the rod is removed?

  1. The sphere becomes negatively charged due to electron flow from ground (correct answer)
  2. The sphere remains positively charged from the original induction
  3. The sphere becomes neutral as all excess charges are removed
  4. The sphere becomes positively charged due to proton transfer from the rod
  5. The charge on the sphere depends on how long grounding is maintained
Explanation: When you encounter charging by induction problems, focus on tracking electron movement and understanding what happens when objects are grounded. Induction involves charge separation without direct contact, and grounding provides a pathway for charge transfer. Let's trace what happens step by step. Initially, the positively charged rod creates an electric field that attracts electrons in the metal sphere toward the rod's side, leaving positive charges on the far side. When you ground the sphere while the rod is still present, you're connecting it to an essentially infinite reservoir of electrons (the Earth). Since the rod is still attracting electrons and creating a positive region on the far side, electrons flow from ground into the sphere to neutralize that positive region. When you remove the rod, those extra electrons remain on the sphere, making it negatively charged. Choice A correctly describes this electron flow from ground, resulting in a negatively charged sphere. Choice B incorrectly assumes the original positive charge remains, but grounding neutralizes this charge and adds excess electrons. Choice C suggests the sphere becomes neutral, which would only happen if you removed the rod before grounding - the order matters crucially in induction. Choice D mentions proton transfer, which is impossible since protons are bound in atomic nuclei and cannot move freely in metals. Remember this key principle: in charging by induction with grounding, the final charge is always opposite to the inducing charge. The timing of when you ground versus when you remove the inducing object determines the final result.

Question 5

A student rubs a glass rod with silk, transferring 1.8×10121.8 \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?

  1. 2.88×10−72.88 \times 10^{-7} C (correct answer)
  2. 1.8×10121.8 \times 10^{12} C
  3. 1.6×10−191.6 \times 10^{-19} C
  4. 2.88×1072.88 \times 10^{7} C
  5. 9.0×10−89.0 \times 10^{-8} C
Explanation: 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×eq = n \times e, where n=1.8×1012n = 1.8 \times 10^{12} electrons and e=1.6×10−19e = 1.6 \times 10^{-19} C per electron. q=(1.8×1012)×(1.6×10−19 C)=2.88×10−7 Cq = (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 (1.8×10121.8 \times 10^{12} C) represents the trap of using the number of electrons as the charge magnitude—forgetting that each electron carries only 1.6×10−191.6 \times 10^{-19} C, not 1 C. This would be an enormous charge in everyday terms. Answer C (1.6×10−191.6 \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 (2.88×1072.88 \times 10^{7} C) results from an incorrect exponent calculation—likely adding exponents incorrectly when multiplying the scientific notation (getting 10−710^{-7} wrong as 10710^{7}). Study tip: Always remember the elementary charge constant e=1.6×10−19e = 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.

Question 6

Two identical conducting spheres are connected by a thin conducting wire. Sphere 1 initially has charge +Q+Q and sphere 2 initially has charge −3Q-3Q. After equilibrium is established through the wire, what is the charge on sphere 1?

  1. −Q-Q (correct answer)
  2. +Q+Q
  3. −2Q-2Q
  4. 00
  5. −Q/2-Q/2
Explanation: 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+Q + (-3Q) = -2Q. After equilibrium, this −2Q-2Q total charge splits equally between the two identical spheres, so each sphere gets −2Q2=−Q\frac{-2Q}{2} = -Q. Therefore, sphere 1 ends up with charge −Q-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+Q, which ignores that charge must redistribute when conductors are connected. Choice (C) gives −2Q-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-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.

Question 7

During a demonstration, a teacher uses a charged ebonite rod to transfer charge to a metal sphere by contact. The rod initially has −15μC-15 \mu\text{C} and after touching the sphere, the rod has −9μC-9 \mu\text{C}. If the sphere was initially neutral, what principle explains why the charges are not equal on the rod and sphere after contact?

  1. The charges distribute according to the capacitances of the objects, not necessarily equally (correct answer)
  2. Conservation of charge is violated during the rapid transfer process
  3. The rod and sphere have different masses, affecting charge distribution
  4. Some charge is lost to the air due to the high voltage involved
  5. The different materials prevent equal charge distribution between objects
Explanation: When you encounter problems involving charge transfer between conductors, the key principle is that charge distributes based on the electrical properties of the objects, particularly their capacitances, not their physical sizes or masses. Let's analyze what happened: The rod lost 6μC6 \mu\text{C} (from −15μC-15 \mu\text{C} to −9μC-9 \mu\text{C}), so the sphere gained +6μC+6 \mu\text{C}. Notice the rod retained more charge than the sphere received. This occurs because when conductors reach electrostatic equilibrium, they have the same electric potential, not the same charge. The relationship Q=CVQ = CV shows that objects with different capacitances (CC) will hold different charges (QQ) at the same potential (VV). Metal spheres typically have different capacitances than ebonite rods due to their geometry and material properties. Choice A correctly identifies that charge distribution depends on capacitance ratios. Choice B is wrong because charge conservation is never violated - the total charge before (−15μC-15 \mu\text{C}) equals the total after (−9μC+(−6μC)=−15μC-9 \mu\text{C} + (-6 \mu\text{C}) = -15 \mu\text{C}). Choice C incorrectly suggests mass affects charge distribution; while mass might influence capacitance indirectly through size, it's not the governing principle. Choice D is incorrect because significant charge loss to air would require much higher voltages than typically achieved with ebonite rods. Remember: In electrostatics problems involving charge sharing, always think about capacitance and potential equilibrium, not equal charge distribution. The objects reach the same potential, with charge distributed according to Q=CVQ = CV.

Question 8

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?

  1. The total charge remained zero throughout the entire process (correct answer)
  2. The total charge became negative while the rod was present
  3. The total charge became positive due to electron repulsion
  4. The total charge oscillated between positive and negative values
  5. The total charge increased due to induction from the external rod
Explanation: 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.

Question 9

Four identical metal spheres W, X, Y, and Z have initial charges of +6q+6q, −2q-2q, +4q+4q, and −8q-8q respectively. Spheres W and X are first brought into contact and separated, then sphere Y is brought into contact with sphere Z and they are separated. Finally, all four spheres are brought together simultaneously. What is the final charge on sphere W?

  1. 00 (correct answer)
  2. +2q+2q
  3. +q+q
  4. −q-q
  5. +4q+4q
Explanation: When identical conducting spheres touch, charge redistributes equally between them. This electrostatics problem tests your understanding of charge conservation and redistribution through a sequence of contacts. Let's trace through each step systematically. Initially: W has +6q+6q, X has −2q-2q, Y has +4q+4q, and Z has −8q-8q. Step 1: W and X touch. Total charge = +6q+(−2q)=+4q+6q + (-2q) = +4q. This splits equally: W gets +2q+2q, X gets +2q+2q. Step 2: Y and Z touch. Total charge = +4q+(−8q)=−4q+4q + (-8q) = -4q. This splits equally: Y gets −2q-2q, Z gets −2q-2q. Step 3: All four spheres touch simultaneously. Total charge = +2q+2q+(−2q)+(−2q)=0+2q + 2q + (-2q) + (-2q) = 0. This distributes equally among four spheres: each gets 00. Therefore, sphere W's final charge is 00. Looking at the wrong answers: Choice (B) +2q+2q represents W's charge after the first contact but ignores the final simultaneous contact. Choice (C) +q+q might result from incorrectly assuming the total charge splits into four equal positive parts. Choice (D) −q-q could come from miscalculating the charge redistribution or incorrectly handling the signs. Study tip: In multi-step electrostatics problems, always track the total charge at each stage—it must be conserved. Work sequentially through each contact, remember that identical spheres always split charge equally, and double-check that your final total charge equals the initial total charge.

Question 10

A student uses a plastic comb to pick up small pieces of paper after running the comb through dry hair. If the comb acquires −2.4×10−9-2.4 \times 10^{-9} C of charge during this process, what was the net change in the number of electrons on the comb?

  1. An increase of 1.5×10101.5 \times 10^{10} electrons (correct answer)
  2. A decrease of 1.5×10101.5 \times 10^{10} electrons
  3. An increase of 2.4×10−92.4 \times 10^{-9} electrons
  4. No change in the number of electrons
  5. A decrease of 3.84×10−283.84 \times 10^{-28} electrons
Explanation: 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=1.6×10−19e = 1.6 \times 10^{-19} C. Since the comb acquired −2.4×10−9-2.4 \times 10^{-9} C, the calculation is: Number of electrons=∣−2.4×10−9 C∣1.6×10−19 C/electron=1.5×1010 electrons\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 1.5×10101.5 \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 2.4×10−92.4 \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 (1.6×10−191.6 \times 10^{-19} C) and that negative charge means excess electrons while positive charge means electron deficiency. The relationship Q=neQ = ne (where nn is the number of electrons) is fundamental to electrostatics problems.

Question 11

An alpha particle (containing 2 protons and 2 neutrons) is emitted from a radioactive nucleus. Before emission, the nucleus had 92 protons. Immediately after the alpha emission, what is the charge of the remaining nucleus in terms of elementary charge units?

  1. +90e+90e (correct answer)
  2. +92e+92e
  3. +94e+94e
  4. +88e+88e
  5. +2e+2e
Explanation: When you encounter nuclear decay problems, focus on conservation laws—specifically, that charge and mass number must be conserved during the process. An alpha particle consists of 2 protons and 2 neutrons, giving it a charge of +2e+2e and a mass number of 4. When this alpha particle is emitted from a nucleus originally containing 92 protons, you need to apply charge conservation. Before decay: nucleus has +92e+92e charge After decay: alpha particle (+2e+2e) + remaining nucleus = +92e+92e total Therefore: remaining nucleus charge = +92e−2e=+90e+92e - 2e = +90e Looking at the wrong answers: Choice B (+92e+92e) represents the misconception that the nucleus charge stays the same—this ignores that the alpha particle carried away charge. Choice C (+94e+94e) suggests incorrectly adding the alpha particle's charge to the original nucleus. Choice D (+88e+88e) involves subtracting 4 instead of 2, likely confusing the alpha particle's mass number (4) with its charge (+2e+2e). The correct answer is A (+90e+90e) because the remaining nucleus has lost exactly 2 protons through alpha emission. Study tip: For any nuclear decay problem, write out the conservation equation explicitly: initial charge = final products' total charge. This systematic approach prevents calculation errors and helps you catch when you're confusing mass number with atomic number. Remember that alpha particles always remove exactly 2 protons and 2 neutrons from the parent nucleus.

Question 12

Two students perform an experiment where they charge identical pith balls by touching them with different charged rods. Ball 1 receives +5.0×10−9+5.0\times 10^{-9} C and ball 2 receives −12.0×10−9-12.0\times 10^{-9} C. The balls are then brought into contact and immediately separated. What fundamental principle determines the final charge distribution?

  1. Conservation of charge requires the total charge before and after contact to be equal (correct answer)
  2. Conservation of energy ensures that the electrostatic energy is minimized
  3. Newton's third law requires equal and opposite charges on the two balls
  4. The uncertainty principle limits how precisely we can know both charges
  5. Coulomb's law determines how the charges redistribute based on distance
Explanation: When you encounter problems involving charged objects coming into contact, you're dealing with electrostatic interactions governed by fundamental conservation laws. The key insight is recognizing which physical principle controls what happens when charges redistribute. Conservation of charge is the fundamental principle at work here. This law states that electric charge can neither be created nor destroyed in an isolated system. When the two pith balls touch, charges redistribute until they reach equilibrium, but the total charge must remain constant. Initially, you have +5.0×10−9+5.0\times 10^{-9} C +(−12.0×10−9)+ (-12.0\times 10^{-9}) C =−7.0×10−9= -7.0\times 10^{-9} C total. After contact, this same total charge will be present, distributed equally between the identical balls (−3.5×10−9-3.5\times 10^{-9} C each). Option A correctly identifies this principle. Option B is incorrect because while energy considerations affect the final equilibrium distribution, conservation of charge is the more fundamental constraint that must be satisfied regardless of energy states. Option C misapplies Newton's third law, which deals with force pairs, not charge distribution—the balls don't need equal and opposite charges. Option D incorrectly invokes quantum mechanics; the uncertainty principle is irrelevant to classical electrostatics problems like this one. Remember: whenever you see charged objects making contact, immediately think "conservation of charge." The total charge before contact always equals the total charge after contact, and for identical conducting objects, this total distributes equally between them.

Question 13

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?

  1. The total charge remains constant as electrons redistribute between components (correct answer)
  2. The total charge decreases as the paint neutralizes against the frame
  3. The total charge increases due to the kinetic energy of the moving droplets
  4. The total charge becomes zero once all paint has been applied
  5. The total charge fluctuates depending on the spray rate and droplet size
Explanation: 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.

Question 14

A technician working with electronic components accidentally touches a charged capacitor plate with a grounded probe. Before contact, the plate had +8.0×10−6+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?

  1. The charge flowed through the probe to ground, conserving total charge in the larger system (correct answer)
  2. The charge was converted to heat energy and destroyed during the discharge process
  3. The charge remained on the plate but became undetectable due to grounding
  4. The charge was neutralized by opposite charges created in the grounding process
  5. The charge was distributed equally between the plate and the probe
Explanation: 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 +8.0×10−6+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.

Question 15

In a photocopying machine, a selenium drum initially has no charge. During operation, it acquires a charge of +4.8×10−9+4.8 \times 10^{-9} C through a charging process. How many electrons were removed from the drum during this charging?

  1. 3.0×10103.0 \times 10^{10} (correct answer)
  2. 4.8×10−94.8 \times 10^{-9}
  3. 7.68×10−287.68 \times 10^{-28}
  4. 1.6×10−191.6 \times 10^{-19}
  5. 1.5×10101.5 \times 10^{10}
Explanation: 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=1.6×10−19e = 1.6 \times 10^{-19} C. Since the drum gained a positive charge of +4.8×10−9+4.8 \times 10^{-9} C, this means electrons (which carry negative charge) were removed. Number of electrons = Total chargeCharge per electron=4.8×10−91.6×10−19=3.0×1010\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 (3.0×10103.0 \times 10^{10}) is correct—this represents the actual number of electrons removed. Choice B (4.8×10−94.8 \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 (7.68×10−287.68 \times 10^{-28}) results from incorrectly multiplying the total charge by the elementary charge instead of dividing: 4.8×10−9×1.6×10−194.8 \times 10^{-9} \times 1.6 \times 10^{-19}. Choice D (1.6×10−191.6 \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=QeN = \frac{Q}{e} where N is the number of electrons, Q is total charge, and e is elementary charge.

Question 16

A Van de Graaff generator transfers electrons from its base to ground, leaving the spherical dome with a positive charge. If 3.2×10143.2 \times 10^{14} electrons are removed from the dome, and each electron carries a charge of magnitude 1.6×10−191.6 \times 10^{-19} C, what is the net charge acquired by the dome?

  1. +5.12×10−5+5.12 \times 10^{-5} C (correct answer)
  2. −5.12×10−5-5.12 \times 10^{-5} C
  3. +3.2×1014+3.2 \times 10^{14} C
  4. +2.0×1033+2.0 \times 10^{33} C
  5. −3.2×1014-3.2 \times 10^{14} C
Explanation: 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 = (3.2×1014)×(1.6×10−19 C)(3.2 \times 10^{14}) \times (1.6 \times 10^{-19} \text{ C}) Total charge = 5.12×10−5 C5.12 \times 10^{-5} \text{ C} Since electrons were removed, the dome becomes positively charged: +5.12×10−5+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 1.6×10−191.6 \times 10^{-19} C is a fundamental constant you should memorize for physics problems.

Question 17

Two identical metal spheres are suspended by insulating threads. Sphere X has a charge of +12μC+12 \mu\text{C} and sphere Y has a charge of −4μC-4 \mu\text{C}. The spheres are brought into contact and then separated. Which statement correctly describes the redistribution of charge?

  1. Each sphere will have a charge of +4μC+4 \mu\text{C} after contact (correct answer)
  2. Sphere X will have +8μC+8 \mu\text{C} and sphere Y will have 0μC0 \mu\text{C}
  3. The charges will completely neutralize, leaving both spheres uncharged
  4. Sphere X will have +6μC+6 \mu\text{C} and sphere Y will have +2μC+2 \mu\text{C}
  5. The final charges cannot be determined without knowing the sphere sizes
Explanation: 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: (+12μC)+(−4μC)=+8μC(+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 +8μC2=+4μC\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 (+8μC+8 \mu\text{C} and 0μC0 \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 (+6μC+6 \mu\text{C} and +2μC+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.

Question 18

Three conducting spheres A, B, and C have charges +8q+8q, +2q+2q, and −4q-4q respectively. All three spheres are simultaneously brought into contact and then separated. What is the final charge on sphere B?

  1. +2q+2q (correct answer)
  2. +6q+6q
  3. +1.5q+1.5q
  4. +3q+3q
  5. +4q+4q
Explanation: When conducting spheres are brought into contact, charge redistributes equally among all spheres until they reach the same potential. This is a fundamental principle of electrostatics that applies whenever conductors touch. To find the final charge on each sphere, you need to apply conservation of charge. First, calculate the total charge in the system: +8q+2q+(−4q)=+6q+8q + 2q + (-4q) = +6q. When all three spheres are brought into simultaneous contact, this total charge distributes equally among them. Since there are three identical conducting spheres, each will have +6q3=+2q\frac{+6q}{3} = +2q after separation. Therefore, sphere B has a final charge of +2q+2q, making choice A correct. Choice B (+6q+6q) represents a common error where students think sphere B somehow acquires all the total charge from the system. Choice C (+1.5q+1.5q) might result from incorrectly calculating the total charge or making an arithmetic error in the division. Choice D (+3q+3q) could come from forgetting that the charges redistribute equally, perhaps thinking each sphere keeps some portion of its original charge plus gains some average amount. The key insight is that identical conducting spheres in contact will always share charge equally, regardless of their initial charge distribution. Remember this rule: when multiple identical conductors touch, add up all the charges and divide by the number of conductors. This principle appears frequently in electrostatics problems and is essential for understanding conductor behavior.

Question 19

In a laboratory demonstration, two conducting spheres are mounted on insulating stands. Sphere A has charge +12μC+12\mu\text{C} and sphere B has charge +3μC+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?

  1. +10μC+10\mu\text{C} (correct answer)
  2. +7.5μC+7.5\mu\text{C}
  3. +6μC+6\mu\text{C}
  4. +9μC+9\mu\text{C}
  5. +5μC+5\mu\text{C}
Explanation: 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=kQRV = \frac{kQ}{R}, where QQ is the charge and RR is the radius. At equilibrium, VA=VBV_A = V_B, so: kQARA=kQBRB\frac{kQ_A}{R_A} = \frac{kQ_B}{R_B} This simplifies to QARA=QBRB\frac{Q_A}{R_A} = \frac{Q_B}{R_B} Since sphere A has twice the radius of sphere B (RA=2RBR_A = 2R_B), we get: QA2RB=QBRB\frac{Q_A}{2R_B} = \frac{Q_B}{R_B} Therefore: QA=2QBQ_A = 2Q_B The total charge is conserved: QA+QB=12μC+3μC=15μCQ_A + Q_B = 12\mu\text{C} + 3\mu\text{C} = 15\mu\text{C} Substituting QA=2QBQ_A = 2Q_B: 2QB+QB=15μC2Q_B + Q_B = 15\mu\text{C} QB=5μCQ_B = 5\mu\text{C} and QA=10μCQ_A = 10\mu\text{C} Answer A (+10μC+10\mu\text{C}) is correct. Answer B (+7.5μC+7.5\mu\text{C}) incorrectly assumes charge splits proportionally by mass or volume. Answer C (+6μC+6\mu\text{C}) wrongly assumes charge distributes inversely with radius. Answer D (+9μC+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.

Question 20

A balloon is rubbed with wool, causing the balloon to become negatively charged with −3.2×10−8-3.2 \times 10^{-8} C. Assuming the wool was initially neutral, how many excess electrons are now on the balloon?

  1. 2.0×10112.0 \times 10^{11} (correct answer)
  2. 3.2×10−83.2 \times 10^{-8}
  3. 5.12×10−275.12 \times 10^{-27}
  4. 1.6×10−191.6 \times 10^{-19}
  5. 1.0×10111.0 \times 10^{11}
Explanation: 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=1.6×10−19e = 1.6 \times 10^{-19} C. Since the balloon has a negative charge of −3.2×10−8-3.2 \times 10^{-8} C, you calculate: Number of electrons = ∣−3.2×10−8∣1.6×10−19=3.2×10−81.6×10−19=2.0×1011\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 (3.2×10−83.2 \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 (5.12×10−275.12 \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 (1.6×10−191.6 \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 1.6×10−191.6 \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.