Redistribution of Charge Between Conductors - College Physics

Identical spheres have $0\,\text{C}$ and $+14\,\mu\text{C}$ ; touch then separate. Final charge on each?
Answer: Each sphere ends with $+7.0\,\mu\text{C}$ This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, charge from charged to neutral is shared equally. Choice A is correct because total +14 μC splits to +7.0 μC each for identicals. Choice C is incorrect as transfer happens. To help students: Treat neutral as zero charge in average. Reinforce with calculations.
Two identical metal rods have $+5\,\text{nC}$ and $-9\,\text{nC}$ . After touching then separating, what is the resulting charge on each?
Answer: Each: $-2\,\text{nC}$ This question tests understanding of charge redistribution between conductors in introductory college physics. Identical rods average charges. Total conserved. Choice A is correct: total -4 nC, each -2. Choice C is incorrect, no change. To help students: Compute net first. Use negatives.
A charged and a neutral identical sphere touch briefly; what happens to the neutral sphere's charge afterward?
Answer: It gains some charge; total charge of both stays constant This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, neutral gains charge from the charged one. Choice B is correct because some charge transfers, total conserved. Choice A is incorrect as neutrals can gain charge. To help students: Think of sharing like dividing resources. Compare with identical charged pairs.
Identical spheres: A has $0\,\text{C}$ , B has $+14\,\text{nC}$ . After touching then separating, what charges result?
Answer: A: $+7\,\text{nC}$ , B: $+7\,\text{nC}$ This question tests understanding of charge redistribution between conductors in introductory college physics. Spheres in contact share charge equally if identical. Equilibrium conserves total charge. Choice A is correct: total +14 nC, each +7 nC. Choice B is incorrect, implying no sharing. To help students: Visualize uncharged gaining charge. Vary initial conditions.
Two identical spheres have $-8.0\,\mu\text{C}$ and $0\,\text{C}$ ; touch then separate. Final charge on each?
Answer: Each sphere ends with $-4.0\,\mu\text{C}$ This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, the identical spheres reach an equilibrium where the negative charge is shared equally. Choice A is correct because the total charge of -8.0 μC is conserved and divided equally, giving -4.0 μC on each. Choice C is incorrect as it implies no charge transfer, ignoring conductor properties. To help students: Visualize negative charge as electrons flowing to equalize. Solve similar problems with positive and negative mixes to avoid sign errors.
Two capacitors in parallel hold $Q_1=+12\,\mu\text{C}$ and $Q_2=+4\,\mu\text{C}$ ; after connecting, total charge is?
Answer: $+16\,\mu\text{C}$ This question tests understanding of charge redistribution between conductors in introductory college physics. When two capacitors are connected in parallel, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, the total charge is the sum before redistribution. Choice A is correct because +12 μC + +4 μC = +16 μC total conserved. Choice D is incorrect as it suggests neutralization. To help students: Remember total Q is invariant. Relate to conductor touching for analogy.
A large conducting sphere of radius R is surrounded by a concentric conducting shell of inner radius 2R and outer radius 3R. The inner sphere carries charge +Q and the shell is initially uncharged. When the shell is connected to ground, what charge appears on the inner surface of the shell?
Answer: The inner surface of the shell acquires charge -Q When you encounter problems involving conducting spheres and shells, think about electrostatic induction and how charges redistribute to maintain zero electric field inside conductors. The key insight is that when the outer shell is grounded, it must develop charges to ensure the electric field inside the conductor material is zero. Since the inner sphere carries charge +Q, electric field lines emanate radially outward from it. These field lines must terminate on negative charges somewhere. In electrostatic equilibrium, all the field lines from the +Q charge on the inner sphere must terminate on the inner surface of the conducting shell. By Gauss's law, if you draw a Gaussian surface inside the shell material (where $\vec{E} = 0$ ), the total enclosed charge must be zero. Since the inner sphere has charge +Q, the inner surface of the shell must have exactly -Q to make the total enclosed charge zero. When the shell is grounded, excess charge can flow to or from ground, but this doesn't change the charge on the inner surface—it only affects the outer surface charge. Answer choice (A) incorrectly assumes no induction occurs. Choice (B) suggests the inner surface gets positive charge, which would violate the requirement that field lines from positive charges terminate on negative charges. Choice (C) gives -Q/2, which might tempt you if you think the charge splits equally, but Gauss's law demands exact cancellation. Remember: in conductor problems, use Gauss's law with surfaces inside the conductor material where $\vec{E} = 0$ to find induced charges. The mathematics forces exact charge cancellation.
Two identical conducting spheres A and B are initially isolated and uncharged. Sphere A is then given a charge of +12 μC. When the spheres are brought into contact and then separated, sphere A is found to have a charge of +3 μC. What was the initial charge on sphere B before contact?
Answer: Sphere B had an initial charge of -6 μC When conducting spheres are brought into contact, charge redistributes until both spheres reach the same potential. For identical spheres, this means equal charge distribution. Let the initial charge on B be $q_B$ . Total charge = 12 + $q_B$ . After contact, each sphere has charge (12 + $q_B$ )/2. Given that A has +3 μC after separation: (12 + $q_B$ )/2 = 3, so 12 + $q_B$ = 6, therefore $q_B$ = -6 μC. Choice A assumes B was uncharged, which would give each sphere +6 μC after contact. Choice C would result in each sphere having +9 μC after contact. Choice D would give each sphere +4.5 μC after contact.
A $+3\,\text{nC}$ rod touches an identical neutral rod, then separates. What is the resulting charge on each rod?
Answer: Each: $+1.5\,\text{nC}$ This question tests understanding of charge redistribution between conductors in introductory college physics. Contact shares with neutral, conserve. Identical equal. Choice A is correct: total +3 nC, each +1.5. Choice B is incorrect, no share. To help students: Include neutrals. Reinforce halves.
Two identical spheres have initial charges $Q_1$ and $Q_2$ and then touch. After separation, what is the final charge on each sphere?
Answer: Each sphere has $(Q_1+Q_2)/2$ due to equal sharing and charge conservation This question tests understanding of charge redistribution between conductors in introductory college physics. Identical spheres share equally. Conservation key. Choice A is correct: each (Q1+Q2)/2. Choice D is incorrect, not neutralizing. To help students: Generalize formulas. Apply to specifics.
A $+9\,\mu\text{C}$ small sphere connects to a larger neutral sphere; after disconnect, which must be true?
Answer: Total remains $+9\,\mu\text{C}$ ; charge splits unequally by size This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, charge from small to larger neutral redistributes unequally. Choice A is correct because total +9 μC remains, split by size. Choice B is incorrect as equal split ignores size. To help students: Note larger capacitance draws more charge. Qualitatively predict for different scenarios.
A small $-10\,\text{nC}$ sphere is connected to a larger neutral sphere by wire. After equilibrium, which statement best describes charge distribution?
Answer: Total stays $-10\,\text{nC}$ , and larger sphere has more negative charge This question tests understanding of charge redistribution between conductors in introductory college physics. Different-sized conductors share charge based on capacity, conserving total. Larger takes more. Choice A is correct: total -10 nC, larger more negative. Choice B is incorrect, assuming equal split. To help students: Think capacitance. Use negative charge examples.
An isolated conducting sphere of radius r carries charge q. A second conducting sphere of radius 3r, initially carrying charge 2q, is brought from infinity and connected to the first sphere by a thin conducting wire. After the system reaches electrostatic equilibrium, what fraction of the total charge resides on the larger sphere?
Answer: The larger sphere contains 3/4 of the total charge When connected, both spheres reach the same potential V = kQ/R. Let q₁ and q₂ be the final charges on spheres of radius r and 3r respectively. For equal potentials: kq₁/r = kq₂/(3r), so q₂ = 3q₁. The total charge is q + 2q = 3q, and charge conservation gives q₁ + q₂ = 3q. Substituting q₂ = 3q₁: q₁ + 3q₁ = 3q, so 4q₁ = 3q, therefore q₁ = 3q/4 and q₂ = 9q/4. The fraction on the larger sphere is q₂/(3q) = (9q/4)/(3q) = 3/4. Choice B would result if charge distributed by surface area ratio. Choice C assumes equal charge distribution. Choice D incorrectly uses the radius cubed ratio.
Identical spheres: A has $-3.0\,\mu\text{C}$ , B has $+1.0\,\mu\text{C}$ . After touching and separating, what charges remain?
Answer: A: $-1.0\,\mu\text{C}$ , B: $-1.0\,\mu\text{C}$ This question tests understanding of charge redistribution between conductors in introductory college physics. Touching enables charge sharing with conservation. Identical spheres share equally. Choice A is correct: total -2.0 μC, each -1.0 μC. Choice B is incorrect, not averaging charges. To help students: Emphasize net charge calculation. Use examples with opposites.
A conducting sphere of radius R carries charge Q. A second uncharged conducting sphere of radius 2R is brought into contact with the first sphere and then separated. What is the ratio of the final charge density on the smaller sphere to the final charge density on the larger sphere?
Answer: The ratio of charge densities is 4:1 When conducting spheres contact, they reach the same potential: V = kq/r. For equilibrium: kq₁/R = kq₂/(2R), so q₂ = 2q₁. Since q₁ + q₂ = Q and q₂ = 2q₁, we get 3q₁ = Q, so q₁ = Q/3 and q₂ = 2Q/3. Surface charge density σ = q/(4πr²). For the small sphere: σ₁ = (Q/3)/(4πR²) = Q/(12πR²). For the large sphere: σ₂ = (2Q/3)/(4π(2R)²) = Q/(24πR²). The ratio σ₁/σ₂ = 4:1. Choice B incorrectly assumes charge distributes by radius ratio. Choice C assumes equal charge densities. Choice D inverts the correct ratio.
Two identical spheres carry $+5\,\text{nC}$ and $-1\,\text{nC}$ . After contact then separation, what is the final charge on each?
Answer: Each: $+2\,\text{nC}$ This question tests understanding of charge redistribution between conductors in introductory college physics. Contact redistributes charge conserving total. Identical spheres equalize charge. Choice A is correct: total +4 nC, each +2 nC. Choice C is incorrect, suggesting neutralization. To help students: Calculate averages first. Explore sign impacts.
Two capacitors connect in parallel; which statement best describes the final charges on them?
Answer: Total charge is conserved; charge redistributes until voltages match This question tests understanding of charge redistribution between conductors in introductory college physics. When two capacitors connect in parallel, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, voltages equalize while total Q is preserved. Choice A is correct because it describes conservation and voltage matching. Choice C is incorrect as charge does transfer. To help students: Recall V = Q/C for equilibrium. Simulate with numbers to see redistribution.
Two capacitors in parallel: $C_1=2\,\mu\text{F}$ with $Q_1=+6\,\mu\text{C}$ , $C_2=1\,\mu\text{F}$ with $Q_2=0$ . After connection, what are $Q_1$ and $Q_2$ ?
Answer: $Q_1=+4\,\mu\text{C}$ , $Q_2=+2\,\mu\text{C}$ This question tests understanding of charge redistribution in capacitors, analogous to conductors in introductory college physics. Parallel connection equalizes voltage, conserving total charge. Charges redistribute per capacitance. Choice A is correct: total 6 μC, V=2V, Q1=4, Q2=2. Choice B is incorrect, equalizing wrongly. To help students: Use Q=CV. Vary capacitances.
A $+7.0\,\mu\text{C}$ rod touches a $-7.0\,\mu\text{C}$ rod; after separation, total charge becomes?
Answer: $0\,\text{C}$ This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, equal opposite charges neutralize overall. Choice C is correct because +7.0 μC + (-7.0 μC) = 0 C total. Choice A is incorrect as it ignores the negative. To help students: Emphasize algebraic summing. Discuss neutralization in equal magnitudes.
A small sphere $+6.0\,\mu\text{C}$ connects by wire to a larger neutral sphere; after disconnect, charges are?
Answer: Total remains $+6.0\,\mu\text{C}$ ; larger holds more charge This question tests understanding of charge redistribution between conductors in introductory college physics. When two conductors come into contact, charge redistributes based on the principle of charge conservation, ensuring the total charge remains constant. In the scenario provided, the different-sized spheres share charge proportional to their capacitances, with the larger one holding more. Choice C is correct because it states the total remains +6.0 μC and the larger sphere gets more charge. Choice B is incorrect as it assumes no transfer to the larger sphere. To help students: Recall that potential equalizes, so Q is proportional to size-related capacitance. Compare with identical cases to highlight differences.

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