3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
Two metal spheres on insulating stands are far apart. Sphere A has radius 0.10 m and charge +7.0 nC; sphere B has radius 0.20 m and charge −1.0 nC. The spheres are then connected by a long, thin wire, which carries a negligible charge, until they reach electrostatic equilibrium. What is then the charge on sphere B?
Answer and reasoning
A+4.0 nCCorrect The total charge is conserved: +7.0 nC − 1.0 nC = +6.0 nC. Connected, the spheres reach the same potential, and far apart each one's potential is kq/R: qA/(0.10 m) = qB/(0.20 m), so qB = 2qA. Then qA + qB = 3qA = 6.0 nC, so qA = +2.0 nC and qB = +4.0 nC.
B+3.0 nC A student who thinks connected conductors share their charge equally picks this: 6.0 nC/2. The spheres end at the same potential, not with the same charge; the larger sphere, with twice the radius, takes twice the charge.
C+4.8 nC A student who thinks the spheres end with the same surface charge density picks this, sharing the charge in proportion to area: 6.0 nC × 0.20²/(0.10² + 0.20²). Equal potentials make the charge proportional to radius, not to area.
D+5.3 nC A student who adds the magnitudes of the charges picks this: 7.0 nC + 1.0 nC = 8.0 nC, shared in proportion to radius, 8.0 × 2/3. With signs, the total is +6.0 nC.
In electrostatics, ground is treated as an idealized reference point. Which statement about ground is correct?
Answer and reasoning
AAny object that is connected to it is left with zero net charge A student who thinks grounding always neutralizes an object picks this. A grounded object is held at zero potential, not zero charge: with a charged object nearby, a grounded conductor keeps a net induced charge.
BIt can take charge away from an object but cannot give any charge to one A student who pictures ground only as a drain for charge picks this. Ground supplies charge as well as absorbing it: electrons flow up from ground onto a grounded conductor near a positive charge.
CIts potential rises a little each time that positive charge flows into it A student who treats ground as an ordinary conductor of limited size picks this. Ground is idealized as able to absorb or provide an infinite amount of charge with no change in its potential, which stays zero.
DIts potential stays zero however much charge flows into or out of itCorrect Ground is an idealized reference point at zero potential that can absorb or provide any amount of charge without its potential changing.
The diagram shows a charged rod held near an uncharged metal sphere on an insulating stand. The sphere is connected to ground by a wire at point F. With the rod still in place, the wire is disconnected; then the rod is taken away. Which describes the sphere's final charge?
Answer and reasoning
ANegative, as electrons came onto the sphere through the wire from ground A student who thinks a charged object induces charge of its own sign picks this. The negative rod repels electrons, so they flow off the sphere to ground, not onto it.
BPositive, since electrons left the sphere through the wireCorrect The negative rod repels free electrons in the sphere, and while the sphere is grounded some of them flow through the wire to ground. Disconnecting the wire traps the deficit, so when the rod is removed the sphere is positive, and its charge spreads over its whole surface.
CPositive, but it all stays on the side where the rod was held A student who thinks induced charge stays where it was induced picks this. Once the rod is taken away, nothing holds the positive charge on one side: it spreads over the whole surface.
DZero, as any object connected to ground is left without charge A student who thinks grounding always leaves an object neutral picks this. While grounded near the negative rod, the sphere is at zero potential with a net positive charge; disconnecting the wire before removing the rod keeps that charge.
Working Rod negative ⇒ electrons repelled through F to ground ⇒ sphere net positive while grounded (V = 0 with rod nearby). Wire removed, then rod removed ⇒ positive charge trapped, spreads uniformly over the surface.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
10.2.A.1 Electrical contact Fix
Electrical contact
Conductors are in electrical contact when they touch or are joined by a conductor such as a wire, so that charge can move from one to the other.
Redistribution of charge
When conductors are put in electrical contact, charge moves between them until their surfaces are all at the same electric potential. The total charge is conserved.
Charge sharing between connected spheres
Two conducting spheres far apart and joined by a thin wire reach one potential: kq₁/R₁ = kq₂/R₂, so the charge divides in proportion to radius. The smaller sphere has the greater surface charge density.
Students often think Conductors that are put in electrical contact always share their total charge equally, whatever their sizes and shapes. In fact No. They end up at the same potential. Only identical conductors share charge equally; two connected spheres far apart share it in proportion to their radii.
Students often think Connected conductors end up with the same surface charge density, so the charge divides in proportion to surface area. In fact No. They end up at the same potential. For two connected spheres far apart, σ is greater on the smaller sphere, since q ∝ R and the area is ∝ R².
10.2.A.2 Ground Fix
Ground
An idealized reference point that has zero electric potential and can absorb or provide an infinite amount of charge without changing its electric potential. Earth is usually treated as ground.
Grounding
Connecting a conductor to ground by a conducting path. The conductor is then held at zero potential, and charge flows to or from ground as needed to keep it there.
Students often think An object connected to ground is always left neutral, because ground cancels any charge. In fact No. A grounded object is held at zero potential, not zero charge. With a charged object nearby, a grounded conductor carries a net induced charge.
Students often think Ground can take charge away from an object but cannot supply charge to it. In fact No. Ground can supply charge as well as absorb it: electrons flow up from ground onto a grounded conductor when that makes its potential zero.
10.2.A.3 Charging by induction Fix
Charging by induction
Giving a conductor a net charge without touching it with a charged object: the conductor is grounded while a charged object is nearby, and the ground connection is broken before the charged object is removed. The conductor is left with charge of the sign opposite to the nearby object's.
Induced charge on a grounded conductor
The net charge that flows onto or off a grounded conductor in an external field; its size is whatever makes the conductor's potential zero, and it is not in general equal and opposite to the external charge.
Students often think Grounding a conductor near a charge q gives the conductor a charge −q, exactly equal and opposite. In fact No. The induced charge is whatever makes the conductor's potential zero. For a grounded sphere of radius R with a point charge q a distance d from its center, it is −qR/d, smaller in magnitude than q.
Students often think Electric potential is calculated with the field expression kq/r² instead of kq/r. In fact No. For a point charge the potential is kq/r, in V; kq/r² is the magnitude of the field, in N/C.
6 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 6
The diagram shows two metal spheres, far apart, connected by a long, thin wire; the distance between them is not drawn to scale. Together they carry a positive charge and are in electrostatic equilibrium. E₁ and E₂ are the magnitudes of the electric field just outside the surfaces of sphere 1 and sphere 2. How does E₁ compare with E₂?
Answer and reasoning
AE₁ is equal to E₂ A student who thinks equal potentials mean equal fields picks this. The spheres are at the same potential V, but the field just outside a sphere is kq/R² = V/R, so for the same V the smaller sphere has the larger field.
BE₁ is one-third of E₂ A student who thinks the sphere with more charge has the stronger field picks this: sphere 2 holds three times the charge. Its surface is also three times as far from its center, so its field is k(3q₁)/(3R)², one-third of E₁.
CE₁ is three times E₂Correct The spheres are at the same potential: kq₁/R = kq₂/(3R), so q₂ = 3q₁. Just outside each surface E = kq/r² with r the sphere's radius: E₁ = kq₁/R² and E₂ = k(3q₁)/(3R)² = kq₁/(3R²). So E₁ = 3E₂: the smaller sphere has the greater surface charge density and the stronger field.
DE₁ is nine times E₂ A student who thinks connected conductors share charge equally picks this: with equal charges, E = kq/R² gives a factor of 3² = 9. The spheres reach the same potential, so q₂ = 3q₁, and E₁ = 3E₂.
An isolated metal sphere of radius R is charged to a potential V₀, taking the potential to be zero far away. It is then connected by a long, thin wire to a distant, uncharged metal sphere of radius 3R. The wire carries a negligible charge. What is the potential of the first sphere once equilibrium is reached?
Answer and reasoning
A0.50V₀ A student who thinks connected conductors share their charge equally picks this: Q/2 left on the first sphere gives k(Q/2)/R = V₀/2. The spheres end at the same potential, which leaves a quarter of the charge on the smaller sphere.
B0.25V₀Correct Initially the sphere's charge is Q = 4πε₀RV₀. Connected, both spheres are at one potential V, with charges 4πε₀RV and 4πε₀(3R)V. Charge is conserved: 4πε₀(4R)V = 4πε₀RV₀, so V = V₀/4 = 0.25V₀.
C0.10V₀ A student who shares the charge in proportion to surface area picks this: R²/(R² + 9R²) = 1/10 of Q on the first sphere, so its potential is V₀/10. The second sphere would then be at k(9Q/10)/(3R) = 0.30V₀, a different potential, so charge would keep flowing.
D0.33V₀ A student who thinks all the charge flows to the larger sphere, as it would to ground, picks this, using kQ/(3R) = V₀/3 as the common potential. Charge flows only until the potentials are equal; the first sphere keeps a quarter of the charge.
Working Q = 4πε₀RV₀. Final common potential V: q₁ = 4πε₀RV, q₂ = 4πε₀(3R)V; q₁ + q₂ = Q ⇒ 4πε₀(4R)V = 4πε₀RV₀ ⇒ V = V₀/4 = 0.25V₀. Distractors: equal shares ⇒ V₀/2; shares ∝ R² ⇒ V₀/10; all charge on the larger sphere ⇒ kQ/(3R) = V₀/3.
A small metal sphere of radius 3.0 cm carries a charge of +9.0 nC. It is lowered through a small hole into a closed, uncharged, hollow metal sphere of radius 6.0 cm on an insulating stand, touched to the inside wall, and then withdrawn without touching anything else. What charge does the small sphere then carry?
Answer and reasoning
A4.5 nC A student who thinks conductors in contact share their charge equally picks this. Equal sharing holds for identical conductors touching side by side; a sphere touching the inside of a hollow conductor gives up all its charge to the outer surface.
B3.0 nC A student who applies the rule for connected spheres far apart picks this: 9.0 nC × 3.0/(3.0 + 6.0). That rule does not apply to a sphere inside a hollow conductor; touching the inside wall, it becomes part of the uncharged cavity wall.
C1.8 nC A student who shares the charge in proportion to surface area picks this: 9.0 nC × 3.0²/(3.0² + 6.0²). The conductors reach one potential, not one charge density, and here that leaves no charge at all on the small sphere.
D0.0 nCCorrect While the small sphere touches the inside wall, it is part of one conductor with the hollow sphere, all at one potential. In equilibrium the excess charge is on the outer surface; with no other charge in the cavity, the cavity wall, and the small sphere touching it, carry no charge. All 9.0 nC moves to the outer surface of the hollow sphere.
Working In contact, small sphere + hollow sphere form one conductor. Excess charge resides on the outer surface; a Gaussian surface in the hollow sphere's metal around the cavity encloses no net charge, and with the cavity otherwise empty the cavity wall (including the small sphere) carries no charge. Small sphere: 0.0 nC; hollow sphere's outer surface: +9.0 nC. Distractors: equal shares 4.5 nC; shares ∝ R: 9.0 × 3.0/9.0 = 3.0 nC; shares ∝ R²: 9.0 × 9.0/45 = 1.8 nC.
A small metal sphere with charge +Q stands on an insulating stand, far from other charges. It is connected to ground by a wire. A student claims that the sphere is then left with no charge. Which reasoning best supports the claim?
Answer and reasoning
AThe sphere is held at zero potential, and far from other charges its potential is zero only if its charge is zero.Correct Connected to ground, the sphere is held at ground's potential, zero. Far from other charges, its potential is set by its own charge alone, V = kQ/R, so V = 0 requires Q = 0: electrons flow up from ground until the sphere is neutral.
BGround cancels any charge, so every object connected to it by a wire is left with zero net charge. A student who thinks grounding always leaves an object neutral picks this. Ground does not cancel charge: it holds a connected conductor at zero potential. Here that means zero charge only because the sphere is far from other charges; a grounded conductor near a charged object keeps a net induced charge.
CCharge flows from a charged object into an uncharged one until the charged object has none left. A student who thinks charge drains from a charged object into an uncharged one picks this. Charge flows only until the connected conductors are at the same potential; between two identical spheres, half the charge would remain.
DThe positive charges on the sphere's surface flow down the wire and into the ground below. A student who thinks positive charges move through metals and wires picks this. In the sphere and the wire only electrons move: the positive sphere becomes neutral because electrons flow up from ground onto it.
The diagram shows a solid metal sphere connected to ground and a fixed point charge near it. The sphere is in electrostatic equilibrium. Which expression gives the net charge on the sphere?
Answer and reasoning
A−q A student who thinks grounding gives a conductor a charge equal and opposite to the nearby charge picks this. The induced charge is whatever makes the sphere's potential zero; its surface is nearer the center (distance R) than +q is (distance d), so a smaller charge, −qR/d, is enough.
B−qR²/d² A student who calculates potential with the field expression kq/r² picks this: kq/d² + kQs/R² = 0. Potential is kq/r; setting kq/d + kQs/R = 0 gives Qs = −qR/d.
C−qR/dCorrect The grounded sphere is at zero potential throughout, including its center. At the center the point charge contributes kq/d, and all the induced charge Qs is on the surface, a distance R from the center, so it contributes kQs/R however it is distributed. kq/d + kQs/R = 0, so Qs = −qR/d.
D0 A student who thinks an object at zero potential has zero charge picks this. The grounded sphere is at zero potential, but that requires a charge whose potential cancels that of +q: electrons flow up from ground, giving the sphere a net charge −qR/d.
Working Grounded ⇒ V = 0 everywhere in the sphere, including its center. Vcenter = kq/d + kQs/R (every element of surface charge is a distance R from the center) = 0 ⇒ Qs = −qR/d. Distractors: −q (equal and opposite); kq/d² + kQs/R² = 0 ⇒ −qR²/d²; V = 0 taken to mean Qs = 0.
Two metal spheres, A of radius 2R and B of radius R, are far apart and initially uncharged. They are connected by a long, thin wire that carries a negligible charge. A point charge +q is then fixed a distance d from the center of sphere B, where d > R; sphere A and the wire are far from the point charge. Take the potential to be zero far from all charges. Once the spheres are in electrostatic equilibrium, which expression gives the charge on sphere A?
Answer and reasoning
A4qR²/(5d²) A student who writes each potential with kq/r² instead of kq/r picks this: qA/(4R²) = −qA/R² + q/d². Those expressions are field magnitudes; potentials are kq/r.
BqR/d A student who treats the distant sphere A as ground picks this: sphere B is then at zero potential, kqB/R + kq/d = 0, so qB = −qR/d and sphere A gains qR/d. Sphere A's potential, kqA/(2R), rises as it gains charge, so it is not at zero potential.
C2qR/(3d)Correct The spheres reach the same potential. Sphere A's potential is kqA/(2R); sphere B's equals the potential at its center, kqB/R + kq/d. With qB = −qA, setting them equal gives qA = 2qR/(3d), the same sign as q.
D0 A student who reasons that the two spheres have zero net charge, so they stay at zero potential and no charge moves, picks this. Sphere B's potential includes kq/d from the point charge, so charge flows along the wire until the two potentials are equal.
Working Connected conductors reach the same potential. Sphere A is far from everything: VA = kqA/(2R). Sphere B is an equipotential, so its potential equals that at its center: every element of its surface charge is a distance R from the center, so its charge contributes kqB/R, and the point charge contributes kq/d: VB = kqB/R + kq/d. Charge conservation: qA + qB = 0. Setting VA = VB: qA/(2R) = −qA/R + q/d, so qA(3/(2R)) = q/d and qA = 2qR/(3d), the same sign as q. Distractors: kq/r² used for potential → qA/(4R²) = −qA/R² + q/d² → qA = 4qR²/(5d²); sphere A treated as ground (sphere B at V = 0) → qB = −qR/d, qA = qR/d; spheres with zero net charge taken to be at zero potential → 0. Checked with sympy (q = 2.0 nC, R = 0.050 m, d = 0.30 m: key 0.22 nC; distractors 0.044 nC, 0.33 nC, 0).
Compiled from the AP Physics C: Electricity and Magnetism Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
