College Physics Quiz: Electrostatics With Conductors

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Electrostatics With ConductorsQuestion 1 of 20

A conducting sphere is placed in a uniform external electric field. After electrostatic equilibrium is established, which statement best describes the electric field inside the conductor?

The field is uniform and parallel to the external field

The field is zero throughout the entire conductor

The field is non-uniform but has the same direction as the external field

The field is uniform but opposite to the external field direction

The field varies linearly from zero at the center to maximum at the surface

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College Physics Quiz

College Physics Quiz: Electrostatics With Conductors

Practice Electrostatics With Conductors 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 Electrostatics With Conductors, 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.

All questions

Question 1

A conducting sphere is placed in a uniform external electric field. After electrostatic equilibrium is established, which statement best describes the electric field inside the conductor?

The field is uniform and parallel to the external field
The field is zero throughout the entire conductor (correct answer)
The field is non-uniform but has the same direction as the external field
The field is uniform but opposite to the external field direction
The field varies linearly from zero at the center to maximum at the surface

Explanation: When you encounter problems involving conductors in electric fields, the key principle is that free charges within conductors can move to establish electrostatic equilibrium. In equilibrium, these mobile electrons redistribute themselves to completely cancel any internal electric field. Here's why the field must be zero inside: If there were any electric field within the conductor, it would exert forces on the free electrons, causing them to move. Since we're told equilibrium is established, no charges are moving, which means no net force acts on them. This is only possible if the electric field is zero throughout the entire conductor's interior. The external field does affect the conductor by inducing charges on its surface—positive charges accumulate where field lines would enter, negative where they would exit—but these surface charges create their own field that exactly cancels the external field inside. Looking at the incorrect options: Choice A is wrong because a uniform internal field parallel to the external field would cause continuous charge movement, preventing equilibrium. Choice C fails for the same reason—any non-zero field, uniform or not, in the same direction as the external field cannot exist in equilibrium. Choice D is also impossible because a uniform field opposite to the external field would still exert forces on charges, causing movement. Remember this fundamental rule: the electric field inside any conductor in electrostatic equilibrium is always zero. This applies regardless of the conductor's shape or the external field configuration—it's a direct consequence of charge mobility within conductors.

Question 2

Two identical conducting spheres, one carrying charge +Q and the other initially uncharged, are brought into contact and then separated. If the spheres are then placed a distance d apart, what is the magnitude of the electrostatic force between them?

$\frac{kQ^2}{d^2}$
$\frac{kQ^2}{4d^2}$ (correct answer)
$\frac{kQ^2}{2d^2}$
$\frac{kQ^2}{8d^2}$
$\frac{kQ^2}{16d^2}$

Explanation: This problem tests your understanding of charge redistribution when conductors come into contact and how to apply Coulomb's law afterward. When two identical conducting spheres touch, charge redistributes equally between them. Since one sphere initially carries charge +Q and the other is uncharged, the total charge is +Q. After contact, each sphere carries +Q/2. This equal distribution occurs because the spheres are identical conductors, and charge flows until electrostatic equilibrium is reached. Once separated by distance d, you can apply Coulomb's law:

F = \frac{kq_1q_2}{d^2}

. Substituting the charges on each sphere:

F = \frac{k(\frac{Q}{2})(\frac{Q}{2})}{d^2} = \frac{kQ^2}{4d^2}

. This confirms answer choice B. Looking at the wrong answers: Choice A (

\frac{kQ^2}{d^2}

) assumes the original charge Q remained on one sphere while the other stayed uncharged—this ignores charge redistribution entirely. Choice C (

\frac{kQ^2}{2d^2}

) represents a common error where students correctly recognize that charge splits but incorrectly apply the factor of 2 only once instead of squaring it in the force calculation. Choice D (

\frac{kQ^2}{8d^2}

) might result from incorrectly assuming unequal charge distribution or mathematical errors in applying the

(\frac{1}{2})^2 = \frac{1}{4}

factor. Remember: When identical conductors touch, charge always distributes equally. In force calculations, both reduced charges matter—don't forget that the force depends on the product of both charges, so the reduction factor gets squared.

Question 3

A conducting shell with inner radius a and outer radius b surrounds a point charge +Q located at the center. What is the charge on the inner surface of the shell?

+Q distributed uniformly over the inner surface
-Q distributed uniformly over the inner surface (correct answer)
Zero, since the conductor shields the inner charge completely
+Q/2 on the inner surface and +Q/2 on the outer surface
-Q/2 on the inner surface and -Q/2 on the outer surface

Explanation: When you encounter electrostatic problems involving conductors, remember that charges in conductors always rearrange themselves to maintain zero electric field inside the conducting material itself. Here's what happens: The point charge +Q at the center creates an electric field that would penetrate into the conductor. However, free electrons in the conductor immediately respond by moving to cancel this field. Since the conductor must have zero field inside its material, the electrons redistribute until they create an exactly opposite field to the central charge. To completely cancel the field from +Q inside the conductor, exactly -Q worth of negative charge must accumulate on the inner surface at radius a. This -Q distributes uniformly over the inner surface because of the spherical symmetry - every point on the inner surface is equidistant from the central charge. Choice A is incorrect because having +Q on the inner surface would double the field inside the conductor rather than cancel it. Choice C misunderstands what "shielding" means - the conductor shields its interior by accumulating charge on surfaces, not by having zero charge. Choice D incorrectly splits the induced charge between surfaces; conservation of charge and Gauss's law require that all -Q sits on the inner surface to cancel the central charge's field within the conductor. Remember: in electrostatics problems with conductors, always apply the principle that the electric field inside conducting material must be zero. This drives all charge redistribution effects.

Question 4

A parallel-plate capacitor has plates separated by distance d and area A. If a conducting slab of thickness t < d is inserted between the plates without touching either plate, how does the capacitance change?

The capacitance decreases because the effective separation increases
The capacitance increases because the effective separation decreases to d - t (correct answer)
The capacitance remains unchanged since the conductor doesn't touch the plates
The capacitance becomes infinite because the conductor shorts the plates
The capacitance increases because the field is concentrated in the gaps

Explanation: When analyzing capacitor problems involving conductors, focus on how the electric field and effective plate separation change. The key insight is that conductors reshape electric field patterns. When you insert a conducting slab between capacitor plates, the conductor becomes an equipotential surface. This effectively creates two capacitors in series: one between the original plate and the conductor's near surface, and another between the conductor's far surface and the other plate. Since the conductor has thickness t, each of these gaps has distance (d-t)/2, making the total effective separation d-t rather than the original distance d. Since capacitance

C = \epsilon_0 A/d_{eff}

, reducing the effective separation from d to (d-t) increases the capacitance. The conductor essentially acts like an additional "floating" plate that concentrates the electric field into smaller gaps. Answer A incorrectly assumes the separation increases - this would only happen if you inserted an insulator that physically pushed the plates apart. Answer C reflects the misconception that the conductor must touch the plates to affect capacitance, but conductors influence electric fields even when isolated. Answer D suggests infinite capacitance from a short circuit, but since the conductor doesn't touch either plate, no direct electrical connection exists. Remember this pattern: inserting any conductor between capacitor plates (without contact) always increases capacitance by reducing effective separation. The conductor redistributes charge and concentrates the electric field, regardless of whether it touches the plates.

Question 5

A conducting sphere of radius R is connected to ground and a point charge +Q is placed at distance d from the center (where d > R). Which statement about the potential at the sphere's surface is correct?

The potential varies over the surface, being most positive nearest to +Q
The potential varies over the surface, being most negative nearest to +Q
The potential is zero everywhere on the surface since the sphere is grounded (correct answer)
The potential equals +kQ/d everywhere on the surface due to the external charge
The potential is undefined because the sphere is grounded and charge flows freely

Explanation: When you encounter problems involving grounded conductors and external charges, the key concept is that grounding maintains a constant potential. A grounded conductor is electrically connected to an infinite reservoir of charge (the Earth), which fixes its potential at zero volts by definition. When the point charge +Q is placed near the grounded conducting sphere, it induces negative charges on the sphere's surface. These induced charges redistribute themselves to ensure that the electric field inside the conductor remains zero. However, regardless of how these charges distribute, the grounding connection ensures that every point on the conductor—including its entire surface—stays at zero potential. The correct answer is C because grounding fundamentally means the potential is fixed at zero everywhere on the conductor. This is true regardless of the external charge configuration. Option A is wrong because while the surface charge density varies (being more negative near +Q), the potential itself cannot vary on a conductor's surface. Option B makes the same error about potential variation, though it correctly notes the surface would be more negative near the positive charge. Option D incorrectly suggests the external charge directly determines the surface potential, ignoring the effect of grounding and the induced charges that develop to maintain zero potential. Remember this key principle: grounding a conductor fixes its potential at zero, period. The conductor will acquire whatever surface charge distribution is needed to maintain this condition, but the potential stays zero everywhere on the conductor.

Question 6

A point charge +q is placed at the center of a conducting spherical shell of inner radius a and outer radius b. The shell is initially uncharged. After equilibrium, what is the electric field at radius r where a < r < b?

$\frac{kq}{r^2}$ directed radially outward
$\frac{kq}{r^2}$ directed radially inward
Zero, because r is inside the conducting material (correct answer)
$\frac{kq}{a^2}$ independent of r
$\frac{kq}{b^2}$ independent of r

Explanation: When you encounter electrostatics problems involving conductors, remember that charges in conductors always rearrange to create zero electric field inside the conducting material itself. Here's what happens: The positive charge +q at the center induces a charge of -q on the inner surface of the conducting shell (at radius a). This induced charge distributes uniformly due to symmetry. To maintain overall neutrality, a charge of +q appears on the outer surface (at radius b). Once equilibrium is reached, the electric field inside any conductor must be zero – this is a fundamental property of conductors in electrostatic equilibrium. Since the region a < r < b lies within the conducting material of the shell, the electric field there is exactly zero. The mobile electrons in the conductor have rearranged themselves to completely cancel any field that the central charge would otherwise create in this region. Option A is wrong because it gives the field you'd expect in empty space due to the central charge, ignoring that we're inside a conductor. Option B has the correct magnitude but wrong direction – though this is irrelevant since the field is actually zero. Option D incorrectly suggests the field depends only on the inner radius and remains constant, which isn't how electrostatics works. Remember this key principle: the electric field is always zero inside a conductor in electrostatic equilibrium, regardless of external charges or induced surface charges. This makes conductor problems much more straightforward once you recognize where the conducting material is located.

Question 7

A conducting wire of length L carries total charge Q distributed over its surface. Assuming the wire is thin compared to its length, where is the charge density highest?

At the center of the wire, where it's farthest from the ends
At the ends of the wire, due to the high curvature there (correct answer)
The charge density is uniform along the entire length
At points one-quarter of the way from each end
The distribution depends on the wire's orientation relative to external fields

Explanation: When you encounter questions about charge distribution on conductors, think about how charges behave to minimize electrostatic energy. Charges on conductors always redistribute to achieve electrostatic equilibrium, and this redistribution is strongly influenced by geometry. For a thin conducting wire, charges accumulate more densely at regions of high curvature because the electric field is stronger there. The ends of the wire represent points of maximum curvature (essentially infinite curvature at the tips), so charges crowd toward these regions to minimize the system's electrostatic energy. This is why lightning rods are pointed – the high charge density at sharp points creates strong electric fields. Choice B correctly identifies that charge density is highest at the wire's ends due to this curvature effect. Choice A suggests the center has the highest charge density, but this contradicts the curvature principle. The center of a wire has minimal curvature, so charges actually avoid this region relative to the ends. Choice C claims uniform distribution, which would only occur for an infinitely long wire or a wire with perfectly rounded ends – neither applies to this thin, finite wire. Choice D proposes maximum density at quarter-points, which has no physical basis. There's nothing special about these locations regarding curvature or electrostatic considerations. Remember this pattern: on any conductor, charge density correlates with curvature. Sharp points and edges accumulate the most charge, while flat surfaces and gentle curves have lower charge density. This principle appears frequently in electrostatics problems.

Question 8

A conducting sphere of radius R is given charge Q. A point P is located at distance r from the center, where r < R (inside the sphere). What is the electric potential at point P?

Zero, because the electric field inside the conductor is zero
$\frac{kQ}{r}$ , same as for a point charge at the center
$\frac{kQ}{R}$ , same as the potential at the surface (correct answer)
$\frac{kQ}{R^2}$ , related to the surface charge density
$\frac{kQ(R-r)}{R^2}$ , accounting for the distance from the surface

Explanation: When dealing with conductors in electrostatic equilibrium, you need to understand two key principles: charge distribution and equipotential surfaces. In a conductor, free electrons redistribute themselves until the electric field inside becomes zero, with all excess charge residing on the surface. The correct answer is C because the entire conductor forms an equipotential surface. Since the electric field inside is zero, no work is required to move a test charge from any interior point to the surface. This means every point inside the conductor, including point P, must have the same potential as the surface. The potential at the surface of a conducting sphere with charge Q is

\frac{kQ}{R}

, so this is also the potential at point P. Answer A makes the common error of confusing electric field with electric potential. While the electric field inside is indeed zero, this doesn't make the potential zero—it makes the potential constant throughout the conductor. Answer B incorrectly treats the situation as if you have a point charge at the center. This ignores the fact that charge resides only on the conductor's surface, not at the center, and fails to account for the conductor's equipotential nature. Answer D has incorrect units and physics. The expression

\frac{kQ}{R^2}

would give units of electric field, not potential, and doesn't represent any meaningful physical quantity for this situation. Remember: conductors in electrostatic equilibrium are always equipotential throughout their volume. When you see conductor problems, immediately think "same potential everywhere inside."

Question 9

A conducting sphere of radius R₁ is placed inside a conducting spherical shell of inner radius R₂ and outer radius R₃, where R₁ < R₂. The inner sphere has charge +Q and the shell has charge +2Q. What is the charge on the outer surface of the shell?

+Q, since charge +Q is induced on the inner surface
+2Q, which is the total charge given to the shell
+3Q, which is the sum of all positive charges in the system (correct answer)
Zero, because all charge resides on the inner surfaces
The charge depends on whether the shell is grounded

Explanation: When you encounter conducting spheres and shells in electrostatics, remember that charges on conductors always redistribute to create zero electric field inside the conductor material. This redistribution follows predictable patterns based on electrostatic induction. Let's trace the charge distribution step by step. The inner sphere carries charge

+Q

. By electrostatic induction, this creates charge

-Q

on the inner surface of the shell (charges of opposite sign are always induced on the nearest conductor surface). Since the shell originally had total charge

+2Q

, and we now have

-Q

on its inner surface, the remaining charge must go somewhere. That "somewhere" is the outer surface of the shell. Using charge conservation: Total charge on shell = Charge on inner surface + Charge on outer surface. Therefore:

+2Q = (-Q) + \text{(charge on outer surface)}

, which gives us

+3Q

on the outer surface. Let's examine why each wrong answer fails. Choice A incorrectly assumes the outer surface charge equals the induced inner surface charge - but these are completely different quantities. Choice B ignores the effect of electrostatic induction entirely, assuming the original shell charge stays put. Choice D misunderstands conductor behavior - charge always moves to outer surfaces when possible, and the shell's outer surface is the most external location available. Study tip: For nested conductors, always apply charge conservation to each conductor separately, accounting for induction effects. The key insight is that induced charges don't disappear - they force redistribution of existing charges.

Question 10

A parallel-plate capacitor has conducting plates of area A separated by distance d. A conducting slab of thickness t and area A is inserted to completely fill the space between the plates. What happens to the capacitance?

The capacitance becomes zero because the plates are effectively shorted
The capacitance becomes infinite because there is no separation between conductors (correct answer)
The capacitance increases by a factor of d/t compared to the original
The capacitance remains $\frac{\varepsilon_0 A}{d}$ because the geometry is unchanged
The capacitance is undefined because the configuration is not valid

Explanation: When analyzing capacitor problems involving conducting materials, you need to understand what happens when conductors are introduced between capacitor plates. A capacitor stores charge by maintaining an electric field between its plates. When you insert a conducting slab that completely fills the space between the plates, you're essentially connecting the two plates with a conductor. In a conductor at electrostatic equilibrium, the electric field inside must be zero, which means there can be no potential difference between any two points within the conductor. Since the conducting slab touches both plates, it forces them to be at the same electric potential. This creates a short circuit between the plates. When there's no potential difference (V = 0) but charge can still flow, the capacitance C = Q/V approaches infinity. This is the correct answer: B. Let's examine why the other options are wrong. Option A incorrectly suggests that shorting the plates makes capacitance zero - this confuses capacitance with resistance. A short circuit has zero resistance but infinite capacitance. Option C attempts to apply the parallel-plate formula incorrectly, treating this as if there were still a dielectric gap of thickness (d-t), but ignores that the conductor creates an equipotential condition. Option D fails to recognize that inserting a conductor fundamentally changes the electrical behavior, not just the geometry. Remember: whenever a conductor completely bridges the gap between capacitor plates, you get a short circuit with infinite capacitance, regardless of the original capacitor geometry.

Question 11

An irregular-shaped conductor in electrostatic equilibrium has regions of different curvature on its surface. Compare the surface charge density σ at a point with small radius of curvature R₁ to that at a point with large radius of curvature R₂ (where R₁ << R₂).

σ₁ = σ₂ because the conductor surface is equipotential
σ₁ < σ₂ because charge spreads more at sharp points
σ₁ > σ₂ because charge concentrates at points of high curvature (correct answer)
The relationship depends on the total charge on the conductor
The relationship depends on the distance between the two surface points

Explanation: When you encounter questions about charge distribution on conductors, focus on the relationship between surface curvature and electric field strength. In electrostatic equilibrium, the electric field just outside a conductor's surface is proportional to the local surface charge density:

E = \frac{\sigma}{\epsilon_0}

. The key insight is that sharper curves (smaller radius of curvature) create stronger electric fields. This happens because field lines must "squeeze together" more tightly around sharp points and protrusions. Since the field strength is directly proportional to surface charge density, regions with high curvature must have higher charge density to produce these stronger fields. Therefore,

\sigma_1 > \sigma_2

when

R_1 << R_2

, making answer C correct. Answer A is incorrect because while the conductor surface is indeed equipotential, this doesn't mean charge density is uniform—it's the potential that's constant, not the charge density. Answer B gets the relationship backwards; charge doesn't "spread more" at sharp points but actually concentrates there. The phrase "spreads more" suggests lower density, which is opposite to reality. Answer D is wrong because the relationship between curvature and charge density is fundamental to electrostatics and doesn't depend on the total charge amount—it's about how that charge distributes itself. Remember this pattern: sharp points and high curvature always mean high charge density and strong electric fields. This principle explains why lightning rods are pointed and why you should avoid sharp metal objects during thunderstorms.

Question 12

An uncharged conducting sphere is placed in the electric field of a point charge +Q located nearby. Which statement correctly describes the force between the point charge and the sphere?

The force is zero because the sphere is initially uncharged
The force is repulsive because positive charge is induced on the sphere
The force is attractive because the induced charges create an attractive interaction (correct answer)
The force alternates between attractive and repulsive as charges redistribute
The force magnitude depends on whether the sphere is grounded or isolated

Explanation: When you encounter problems involving conductors in electric fields, remember that free charges in conductors always rearrange themselves to reach electrostatic equilibrium, and this redistribution creates forces even when the conductor starts uncharged. Here's what happens: When the positive point charge +Q is placed near the uncharged conducting sphere, the free electrons in the conductor are attracted toward +Q and accumulate on the side of the sphere closest to the point charge. This leaves a deficit of electrons (net positive charge) on the far side of the sphere. The sphere now has induced negative charge on the near side and induced positive charge on the far side. Since the negative charges are closer to +Q than the positive charges, the attractive force between +Q and the nearby negative charges is stronger than the repulsive force between +Q and the distant positive charges. The net result is always an attractive force between the point charge and the conductor. Option A is incorrect because even though the sphere has zero net charge, the redistribution of that charge creates local regions of positive and negative charge that interact with the external field. Option B misses that while positive charge is indeed induced, it's on the far side of the sphere, and the dominant interaction comes from the closer negative charges. Option D is wrong because once equilibrium is reached, the force remains constant and attractive. Remember: conductors in external electric fields always experience attractive forces due to charge redistribution, regardless of the conductor's initial charge state.

Question 13

An isolated conducting sphere of radius R carries total charge Q. A small test charge +q is placed at distance r from the center (where r > R). What is the magnitude of the electric field at the test charge location?

$\frac{kQ}{R^2}$
$\frac{kQ}{r^2}$ (correct answer)
$\frac{kQ}{(r-R)^2}$
$\frac{kQ}{4\pi r^2}$
$\frac{kQq}{r^2}$

Explanation: When you encounter problems involving conducting spheres and electric fields, remember that charge distribution on conductors follows specific rules that simplify the analysis significantly. For an isolated conducting sphere carrying charge Q, all the charge resides on the outer surface due to electrostatic repulsion. This creates a key insight: from any external point, the sphere behaves exactly like a point charge Q located at its center. The sphere's radius R becomes irrelevant for calculating the external electric field. Using Coulomb's law for the electric field created by a point charge, the magnitude at distance r from the center is

E = \frac{kQ}{r^2}

, making choice B correct. Let's examine why the other options are wrong. Choice A (

\frac{kQ}{R^2}

) incorrectly uses the sphere's radius instead of the distance to the test charge - this would give the field strength at the surface, not at the test charge location. Choice C (

\frac{kQ}{(r-R)^2}

) represents a common misconception that you should measure distance from the surface rather than the center - but the sphere acts as a point charge at its center, so the full distance r matters. Choice D (

\frac{kQ}{4\pi r^2}

) omits the proportionality constant k, making it dimensionally incorrect for electric field. Study tip: Whenever you see a conducting sphere problem, immediately think "point charge at the center" for external field calculations. The conductor's size only matters for points inside or on the surface.

Question 14

Two conducting spheres of radii R₁ and R₂ are connected by a thin conducting wire. If the system carries total charge Q, what is the ratio of surface charge densities σ₁/σ₂?

R₁/R₂
R₂/R₁ (correct answer)
(R₁/R₂)²
(R₂/R₁)²
1 (the charge densities are equal)

Explanation: When conductors are connected electrically, they form an equipotential system - meaning they must have the same electric potential everywhere. This fundamental principle is key to solving problems involving connected conducting objects. For a conducting sphere with charge Q and radius R, the potential is

V = \frac{kQ}{R}

and the surface charge density is

\sigma = \frac{Q}{4\pi R^2}

. Since our two spheres are connected by a wire, they must have equal potentials:

V_1 = V_2

, which gives us

\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2}

. This simplifies to

\frac{Q_1}{R_1} = \frac{Q_2}{R_2}

, or

Q_1 = Q_2 \cdot \frac{R_1}{R_2}

. Now we can find the charge density ratio. Since

\sigma_1 = \frac{Q_1}{4\pi R_1^2}

and

\sigma_2 = \frac{Q_2}{4\pi R_2^2}

, we get:

\frac{\sigma_1}{\sigma_2} = \frac{Q_1 R_2^2}{Q_2 R_1^2} = \frac{Q_2 \cdot \frac{R_1}{R_2} \cdot R_2^2}{Q_2 R_1^2} = \frac{R_2}{R_1}

This confirms answer (B) is correct. (A) R₁/R₂ incorrectly inverts the relationship. (C) (R₁/R₂)² might arise from confusing this with capacitance relationships or incorrectly handling the surface area terms. (D) (R₂/R₁)² makes the same error as (C) but in the opposite direction. Remember: smaller conducting spheres always have higher surface charge density when connected to larger ones. The inverse relationship

\sigma \propto \frac{1}{R}

is a direct consequence of the equipotential condition.

Question 15

A conducting cube is placed in a uniform electric field. In electrostatic equilibrium, where is the charge density greatest on the cube's surface?

At the centers of the faces, where the surface is flattest
At the edges, where two faces meet
At the corners, where three faces meet (correct answer)
The charge density is uniform over the entire surface
At the faces perpendicular to the external field direction

Explanation: When a conductor is placed in an electric field, you're dealing with electrostatic equilibrium and surface charge distribution. The key principle here is that electric field lines must be perpendicular to the conductor's surface, and the field strength (and therefore charge density) depends on the surface's curvature. In electrostatic equilibrium, charges redistribute on the conductor's surface to create zero electric field inside. The surface charge density

\sigma

is directly proportional to the local electric field strength just outside the surface. Sharp points and edges create stronger electric fields because they have higher curvature, concentrating the field lines. The corners of a cube represent the highest curvature points where three faces meet at right angles. This extreme geometry creates the strongest electric field concentration, resulting in the highest charge density. Answer C is correct. Answer A is wrong because flat face centers have minimal curvature, producing the weakest fields and lowest charge densities. Answer B is incorrect because edges have moderate curvature - more than faces but less than corners, so they have intermediate charge density. Answer D represents a common misconception; uniform charge distribution would only occur on a sphere, where all points have identical curvature. Remember this pattern: on any conductor surface, charge density increases with curvature. Sharp points have the highest density, flat areas the lowest. This principle explains why lightning rods work and why you should avoid sharp objects during thunderstorms - the field concentration at sharp points is enormous.

Question 16

A point charge +q is placed near a grounded conducting sphere. Which statement correctly describes the induced charge distribution on the sphere's surface?

Positive charge is induced uniformly over the entire surface of the conductor
Negative charge is induced uniformly over the entire surface of the conductor
Negative charge is induced on the surface closest to +q, positive charge on the far surface
The total induced charge is zero, with negative charge concentrated near +q (correct answer)
No charge is induced since the conductor is grounded and electrically neutral

Explanation: When you encounter problems involving conductors and external charges, remember that conductors have free electrons that can move to reach electrostatic equilibrium. The key principle is that the electric field inside a conductor must be zero. When the positive charge +q approaches the grounded conducting sphere, it attracts the free electrons in the conductor toward the near surface, leaving positive charges on the far surface. However, since the sphere is grounded, it's connected to an infinite reservoir of charge (typically Earth). Electrons flow from ground onto the sphere until the electric field inside becomes zero. This process continues until enough negative charge accumulates to completely cancel the field from +q inside the conductor. The correct answer is D because grounding ensures the total induced charge equals

-q

(to maintain zero field inside), and this negative charge concentrates nearest to +q where the attractive force is strongest. The charge distribution is non-uniform, with higher density closer to the external charge. Answer A is wrong because grounding provides electrons, making the net charge negative, not positive. Answer B fails because the charge distribution cannot be uniform—the external charge creates a non-uniform electric field that requires a non-uniform surface charge to cancel it inside the conductor. Answer C incorrectly suggests positive charge remains on the far surface; grounding allows enough electrons to flow onto the sphere to make the entire surface negatively charged. Remember: grounded conductors always acquire whatever total charge is needed to zero the internal field, and this charge distributes non-uniformly based on the external field geometry.

Question 17

Two identical conducting spheres A and B are initially separated by a large distance. Sphere A carries charge $+3Q$ and sphere B carries charge $-Q$ . The spheres are brought into contact and then separated to their original positions. What is the final charge on each sphere?

Sphere A: $+2Q$ , Sphere B: $0$
Sphere A: $+Q$ , Sphere B: $+Q$ (correct answer)
Sphere A: $+3Q$ , Sphere B: $-Q$
Sphere A: $+1.5Q$ , Sphere B: $+0.5Q$

Explanation: When identical conductors are brought into contact, charge redistributes until both conductors reach the same potential. For identical spheres, this means equal charge distribution. Total initial charge is

+3Q + (-Q) = +2Q

. This total charge divides equally between the two identical spheres:

+2Q/2 = +Q

on each sphere.

Question 18

A hollow conducting sphere with inner radius $a$ and outer radius $b$ has a point charge $+Q$ placed at its center. If the sphere is initially uncharged, what is the surface charge density on the outer surface?

$\sigma = \frac{Q}{4\pi a^2}$
$\sigma = \frac{Q}{4\pi b^2}$ (correct answer)
$\sigma = \frac{Q}{4\pi(b^2 - a^2)}$
$\sigma = \frac{Q}{8\pi b^2}$

Explanation: By Gauss's law and the properties of conductors, the point charge

+Q

at the center induces a total charge

-Q

uniformly distributed on the inner surface and

+Q

uniformly distributed on the outer surface (to maintain overall neutrality). The outer surface area is

4\pi b^2

, so the surface charge density is

\sigma = \frac{Q}{4\pi b^2}

Question 19

A conducting sphere of radius $R$ is given a charge $Q$ . A second, smaller uncharged conducting sphere of radius $r$ (where $r < R$ ) is brought into contact with the first sphere and then removed. What fraction of the original charge remains on the large sphere?

$\frac{R}{R + r}$ (correct answer)
$\frac{R^2}{R^2 + r^2}$
$\frac{R^3}{R^3 + r^3}$
$\frac{R - r}{R + r}$

Explanation: When conducting spheres are in contact, they reach the same potential. For a conducting sphere,

V = k\frac{Q}{R}

. If the large sphere has charge

Q_1

and small sphere has charge

Q_2

after contact, then

\frac{kQ_1}{R} = \frac{kQ_2}{r}

, giving

\frac{Q_1}{Q_2} = \frac{R}{r}

. Since

Q_1 + Q_2 = Q

, we get

Q_1 = \frac{RQ}{R + r}

. The fraction remaining on the large sphere is

\frac{Q_1}{Q} = \frac{R}{R + r}

Question 20

Two conducting spheres of radii $R_1 = 2R$ and $R_2 = R$ are connected by a thin conducting wire and carry a total charge $Q$ . What is the ratio of the surface charge densities $\frac{\sigma_1}{\sigma_2}$ ?

$\frac{1}{4}$
$2$
$1$
$\frac{1}{2}$ (correct answer)

Explanation: When two conducting spheres are connected by a wire, they form an equipotential system. This means both spheres must have the same electric potential, which is the key insight for solving this problem. For a conducting sphere with charge

Q

and radius

R

, the potential is

V = \frac{kQ}{R}

and the surface charge density is

\sigma = \frac{Q}{4\pi R^2}

. Since the spheres are at equal potential:

\frac{kQ_1}{R_1} = \frac{kQ_2}{R_2}

, which gives us

\frac{Q_1}{Q_2} = \frac{R_1}{R_2} = \frac{2R}{R} = 2

. With the constraint

Q_1 + Q_2 = Q

and

Q_1 = 2Q_2

, we find

Q_2 = \frac{Q}{3}

and

Q_1 = \frac{2Q}{3}

. Now we can find the charge densities:

\sigma_1 = \frac{Q_1}{4\pi R_1^2} = \frac{2Q/3}{4\pi(2R)^2} = \frac{Q}{24\pi R^2}

and

\sigma_2 = \frac{Q_2}{4\pi R_2^2} = \frac{Q/3}{4\pi R^2} = \frac{Q}{12\pi R^2}

. Therefore:

\frac{\sigma_1}{\sigma_2} = \frac{Q/(24\pi R^2)}{Q/(12\pi R^2)} = \frac{1}{2}

, confirming answer D. The wrong answers represent common mistakes: A)

\frac{1}{4}

incorrectly uses the area ratio; B)

2

confuses the charge ratio with the density ratio; C)

1

assumes equal charge densities, ignoring the size difference. Remember: connected conductors have equal potentials, not equal charge densities. The smaller sphere always has higher charge density.