College Physics Quiz: Electric Fields Of Charge Distributions

20 questions · exam conditions

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Electric Fields Of Charge DistributionsQuestion 1 of 20

An infinite plane sheet carries uniform surface charge density $+\sigma$ . A second identical sheet is placed parallel to the first at distance $d$ . What is the electric field magnitude in the region between the two sheets?

0

\frac{\sigma}{4\epsilon_0}

\frac{\sigma}{2\epsilon_0}

\frac{\sigma}{\epsilon_0}

\frac{2\sigma}{\epsilon_0}

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College Physics Quiz

College Physics Quiz: Electric Fields Of Charge Distributions

Practice Electric Fields Of Charge Distributions 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 Electric Fields Of Charge Distributions, 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

$0$
$\frac{\sigma}{4\epsilon_0}$
$\frac{\sigma}{2\epsilon_0}$
$\frac{\sigma}{\epsilon_0}$ (correct answer)
$\frac{2\sigma}{\epsilon_0}$

Explanation: When you encounter problems with charged sheets, think about electric field superposition. Each infinite sheet with surface charge density

+\sigma

creates its own uniform electric field of magnitude

\frac{\sigma}{2\epsilon_0}

pointing away from the sheet (since both sheets are positively charged). To find the total field between the sheets, you need to add the contributions from both sheets as vectors. Consider a point between the two sheets. The left sheet creates a field pointing rightward (away from it) with magnitude

\frac{\sigma}{2\epsilon_0}

. The right sheet also creates a field pointing rightward (toward the left, away from the right sheet) with the same magnitude

\frac{\sigma}{2\epsilon_0}

. Since both fields point in the same direction between the sheets, they add:

\frac{\sigma}{2\epsilon_0} + \frac{\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}

. Option A (

0

) would only occur if the sheets had opposite charges, causing their fields to cancel between them. Option B (

\frac{\sigma}{4\epsilon_0}

) has no physical basis for this configuration. Option C (

\frac{\sigma}{2\epsilon_0}

) represents the field from just one sheet, ignoring the contribution from the second sheet entirely. The correct answer is D:

\frac{\sigma}{\epsilon_0}

. Study tip: For parallel charged sheets, always sketch the field directions from each sheet separately, then use vector addition. Remember that fields add between like-charged sheets but cancel between oppositely-charged sheets.

Question 2

A solid conducting sphere of radius $R$ carries total charge $Q$ . Using Gauss's law, what can be concluded about the electric field inside the conductor ( $r < R$ )?

The field is uniform and equals $\frac{Q}{4\pi\epsilon_0 R^2}$
The field varies as $\frac{1}{r^2}$ and equals $\frac{Q}{4\pi\epsilon_0 r^2}$
The field is zero everywhere inside the conductor (correct answer)
The field is maximum at the center and decreases toward the surface
Gauss's law cannot determine the field inside a conductor

Explanation: When you encounter problems involving conductors and electric fields, the key principle is that charges in conductors always redistribute to create electrostatic equilibrium. This means no electric field can exist inside the conducting material itself. Here's why the field must be zero everywhere inside: In electrostatic equilibrium, free charges in the conductor have stopped moving. If there were any electric field inside the conductor, it would exert forces on these free charges, causing them to move until they arranged themselves to cancel that field completely. This redistribution continues until the net electric field inside becomes exactly zero. Using Gauss's law confirms this: if you draw any Gaussian surface entirely within the conductor, it must enclose zero net charge (since all charge resides on the surface), so the electric field through that surface must be zero. Looking at the wrong answers: Option A suggests a uniform field, but this would cause charge movement, violating equilibrium. Option B applies the field equation for a point charge, which ignores the conductor's fundamental property of charge redistribution. Option D proposes a varying field inside, but again, any internal field would drive charge motion until it's eliminated. The correct answer is C - the field is zero everywhere inside the conductor. Study tip: Remember that "inside a conductor" always means zero electric field in electrostatics problems. The charges live only on the surface, and they arrange themselves precisely to cancel any internal field. This is true regardless of the conductor's shape or total charge.

Question 3

Two point charges, $+2q$ and $-q$ , are separated by distance $d$ . At what point along the line connecting them is the electric field zero?

At distance $\frac{d}{3}$ from $+2q$
At distance $\frac{d}{2}$ from $+2q$
At distance $\frac{2d}{3}$ from $+2q$
At distance $d$ from $+2q$ (beyond $-q$ )
At distance $2d$ from $+2q$ (beyond $-q$ ) (correct answer)

Explanation: When analyzing electric fields from multiple point charges, you need to find where the individual electric field vectors cancel each other out. This requires understanding both the magnitude and direction of electric fields. The electric field from a point charge has magnitude

E = k\frac{|q|}{r^2}

and points away from positive charges, toward negative charges. For the total field to be zero, the fields from both charges must have equal magnitudes but opposite directions. Let's say the zero-field point is distance

x

from the

+2q

charge, making it distance

(d-x)

from the

-q

charge. Both fields must point in the same direction along the line for them to cancel - this only happens between the two charges, where the

+2q

field points away from it (rightward) and the

-q

field points toward it (also rightward). Setting the magnitudes equal:

k\frac{2q}{x^2} = k\frac{q}{(d-x)^2}

Simplifying:

\frac{2}{x^2} = \frac{1}{(d-x)^2}

Cross-multiplying:

2(d-x)^2 = x^2

Taking square roots:

\sqrt{2}(d-x) = x

Solving:

x = \frac{d}{\sqrt{2}+1} = \frac{d(\sqrt{2}-1)}{(\sqrt{2}+1)(\sqrt{2}-1)} = d(\sqrt{2}-1) \approx 0.414d

This doesn't match any given option exactly, suggesting there may be an error in the problem or answer choices. Options A, B, and C (

\frac{d}{3}

\frac{d}{2}

\frac{2d}{3}

) are all between the charges but give incorrect distances. Option D places the point beyond the

-q

charge, where the fields would add rather than cancel. Strategy tip: Always check that electric field vectors point in opposite directions at the zero-field point - this geometric constraint often eliminates several answer choices immediately.

Question 4

A spherical shell of radius $R$ carries total charge $Q$ distributed uniformly over its surface. What is the electric field at distance $r = \frac{R}{2}$ from the center?

$\frac{Q}{4\pi\epsilon_0 R^2}$
$\frac{Q}{2\pi\epsilon_0 R^2}$
$\frac{Q}{\pi\epsilon_0 R^2}$
$\frac{2Q}{\pi\epsilon_0 R^2}$
$0$ (correct answer)

Explanation: When you encounter problems involving charged spherical shells, immediately think about Gauss's law and the symmetry of the situation. The key insight is that electric field depends on where you're measuring it relative to the charge distribution. For a uniformly charged spherical shell, you need to consider two regions: inside the shell (r < R) and outside the shell (r > R). Since you're asked about the field at

r = \frac{R}{2}

, which is inside the shell, this is the crucial case to understand. Using Gauss's law with a spherical Gaussian surface of radius

r = \frac{R}{2}

, you can see that this surface encloses zero charge because all the charge

Q

resides on the shell at radius

R

. Since the enclosed charge is zero, Gauss's law gives us

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0} = 0

. By symmetry, the electric field must be constant over the Gaussian surface, so

E = 0

everywhere inside the shell. The answer choices A through D all give non-zero values, representing common misconceptions. Choice A,

\frac{Q}{4\pi\epsilon_0 R^2}

, is what you'd get if you incorrectly treated the shell like a point charge at the center. Choices B, C, and D involve similar errors with different geometric factors, perhaps from misapplying formulas or incorrectly accounting for the surface charge distribution. Remember this fundamental result: the electric field inside any uniformly charged spherical shell is always zero, regardless of the shell's charge or radius. This is a direct consequence of Gauss's law and spherical symmetry.

Question 5

A thin semicircular rod of radius $R$ carries uniform linear charge density $\lambda$ . What is the direction of the electric field at the center of the semicircle?

Along the axis of symmetry, toward the curved part
Along the axis of symmetry, away from the curved part (correct answer)
Perpendicular to the axis of symmetry
At 45° to the axis of symmetry
The field is zero by symmetry

Explanation: When analyzing electric fields from charged objects, symmetry is your most powerful tool. For a uniformly charged semicircular rod, you need to consider how charge elements contribute to the net field at the center. Imagine breaking the semicircle into tiny charge elements. Each element

dq = \lambda \, dl

creates an electric field pointing radially outward from that element toward the center (assuming positive charge). Due to the semicircle's symmetry about its central axis, all horizontal components of these individual electric field vectors cancel out perfectly. The charge element at the far left creates a field component pointing right, which exactly cancels the leftward component from the element at the far right. However, all the vertical components point in the same direction - away from the curved part of the semicircle, along the axis of symmetry. These components add constructively, creating a net field pointing away from the rod. Choice A incorrectly suggests the field points toward the curved part, which would only happen with negative charge. Choice C claims the field is perpendicular to the symmetry axis, but symmetry arguments show the horizontal components must cancel. Choice D suggests a 45° angle, which has no physical basis given the semicircle's geometry. The correct answer is B - the field points along the axis of symmetry, away from the curved part. Key strategy: In symmetric charge distributions, always identify which field components cancel due to symmetry and which components reinforce each other. This approach works for rings, arcs, and other symmetric geometries.

Question 6

A conducting spherical shell has inner radius $a$ and outer radius $b$ . A point charge $+Q$ is placed at the center. What is the surface charge density on the outer surface of the conductor?

$\frac{Q}{4\pi a^2}$
$\frac{Q}{4\pi b^2}$ (correct answer)
$\frac{Q}{4\pi(b^2 - a^2)}$
$\frac{Q}{2\pi b^2}$
$0$

Explanation: When you encounter a conductor with charges, remember that electrostatic equilibrium creates a predictable charge distribution. The key principle is that charges in a conductor always reside on surfaces, and the electric field inside the conductor must be zero. With a point charge

+Q

at the center of this spherical shell, electrostatic induction occurs. The conductor responds by placing exactly

-Q

on its inner surface to shield the conducting material from the electric field. Since charge is conserved and the conductor was initially neutral, the outer surface must have

+Q

to balance the

-Q

on the inner surface. The crucial insight is that this

+Q

charge distributes uniformly over the outer surface due to spherical symmetry. Surface charge density is simply total charge divided by surface area, so:

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

, making choice B correct. Choice A incorrectly uses the inner radius

a

, but the charge we're analyzing sits on the outer surface at radius

b

. Choice C uses the difference

b^2 - a^2

, which might seem logical if you mistakenly think about the "volume" of the shell, but surface charge density depends only on the surface area where the charge actually resides. Choice D has the wrong geometric factor—it uses

2\pi

instead of

4\pi

, confusing the area of a circle with that of a sphere. Remember: in conductor problems, always identify where charges end up (surfaces only), then use symmetry to find how they distribute. The geometry of the specific surface determines the area calculation.

Question 7

A Gaussian surface in the shape of a cylinder with radius $R$ and length $L$ is placed with its axis parallel to a uniform electric field $\vec{E} = E_0\hat{i}$ . What is the total electric flux through the cylindrical surface?

$E_0 \pi R^2 L$
$2E_0 \pi R^2$
$E_0 \pi R^2$
$E_0 \cdot 2\pi R L$
$0$ (correct answer)

Explanation: When working with Gauss's law, you need to calculate the electric flux through each part of a closed surface separately, then sum them up. For a cylinder, this means considering the two circular end caps and the curved side surface. The electric flux through any surface is

\Phi = \vec{E} \cdot \vec{A} = EA\cos\theta

, where

\theta

is the angle between the electric field and the surface normal vector. For the cylinder's end caps: Each circular end has area

\pi R^2

. Since the field

\vec{E} = E_0\hat{i}

is parallel to the cylinder's axis, it's either parallel or antiparallel to each end cap's normal vector. One end contributes flux

+E_0\pi R^2

and the other contributes

-E_0\pi R^2

. These cancel out completely. For the curved side surface: The electric field is parallel to the cylinder axis, while the normal vectors on the curved surface point radially outward (perpendicular to the axis). Since

\theta = 90°

everywhere on the curved surface,

\cos\theta = 0

, giving zero flux contribution. The total flux is therefore zero:

0 + 0 + 0 = 0

. Looking at the wrong answers: A)

E_0 \pi R^2 L

incorrectly treats the entire cylinder volume as contributing flux. B)

2E_0 \pi R^2

counts both end caps as positive flux, ignoring their opposite orientations. C)

E_0 \pi R^2

counts only one end cap. D)

E_0 \cdot 2\pi R L

incorrectly assumes the curved surface contributes flux. Remember: when a uniform field is parallel to a cylinder's axis, the net flux through the entire closed surface is always zero due to symmetry.

Question 8

Two parallel infinite planes carry surface charge densities $+\sigma$ and $-\sigma$ respectively, separated by distance $d$ . A small test charge $+q$ is moved from the negative plane to the positive plane. What is the work done by the electric field?

$\frac{\sigma q d}{\epsilon_0}$
$\frac{\sigma q d}{2\epsilon_0}$
$-\frac{\sigma q d}{\epsilon_0}$ (correct answer)
$-\frac{\sigma q d}{2\epsilon_0}$
$0$

Explanation: When dealing with parallel charged planes, you need to understand how electric fields superpose and how work relates to electric potential difference. Each infinite plane creates a uniform electric field of magnitude

\frac{\sigma}{2\epsilon_0}

, and the total field between oppositely charged planes adds up to

\frac{\sigma}{\epsilon_0}

pointing from positive to negative. The work done by the electric field equals

W = q \cdot E \cdot d \cdot \cos\theta

, where

\theta

is the angle between field direction and displacement. Since you're moving the positive test charge from the negative plane to the positive plane, you're moving against the electric field direction (the field points from + to -, but you're going from - to +). This makes

\theta = 180°

, so

\cos\theta = -1

. Therefore:

W = q \cdot \frac{\sigma}{\epsilon_0} \cdot d \cdot (-1) = -\frac{\sigma q d}{\epsilon_0}

The negative sign confirms that the electric field does negative work, which makes physical sense since you're moving a positive charge against the field. Looking at the wrong answers: (A)

\frac{\sigma q d}{\epsilon_0}

ignores the direction and gives positive work. (B)

\frac{\sigma q d}{2\epsilon_0}

uses only one plane's field strength and wrong sign. (D)

-\frac{\sigma q d}{2\epsilon_0}

has the right sign but uses only one plane's contribution instead of the combined field. Study tip: For parallel plate problems, remember that the field between oppositely charged plates is

\frac{\sigma}{\epsilon_0}

, and work done by a field is negative when moving against it.

Question 9

A point charge $+q$ is located at the center of a spherical Gaussian surface of radius $r$ . If the radius is doubled while keeping the charge fixed, what happens to the electric flux through the surface and the electric field at the surface?

Flux decreases by factor of 4; field decreases by factor of 4
Flux decreases by factor of 2; field decreases by factor of 4
Flux remains constant; field decreases by factor of 4 (correct answer)
Flux remains constant; field decreases by factor of 2
Flux increases by factor of 4; field remains constant

Explanation: This question tests two fundamental electrostatics concepts: Gauss's law for electric flux and Coulomb's law for electric field strength. When you encounter Gaussian surface problems, always consider these principles separately. Electric flux depends only on the enclosed charge according to Gauss's law:

\Phi = \frac{q_{enc}}{\epsilon_0}

. Since the charge

+q

remains fixed at the center regardless of the sphere's radius, the total flux through any spherical surface surrounding it stays constant. The flux doesn't depend on the surface's size or distance from the charge. Electric field follows Coulomb's law:

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

, where

r

is the distance from the point charge. When you double the radius from

r

2r

, the field becomes

E = \frac{kq}{(2r)^2} = \frac{kq}{4r^2}

, which is one-fourth the original value. Looking at the wrong answers: Choice A incorrectly assumes flux depends on surface area, confusing flux with field calculations. Choice B makes the same flux error while getting the field calculation right. Choice D correctly identifies that flux remains constant but incorrectly applies an inverse relationship (factor of 2) instead of the inverse-square relationship for electric field. Study tip: Remember that Gauss's law makes flux independent of surface geometry when the enclosed charge is unchanged, while electric field always follows the inverse-square law with distance. Practice distinguishing between these two concepts—they're tested together frequently but follow completely different rules.

Question 10

A thin ring of radius $R$ carries uniform charge $Q$ . At what distance along the axis from the center does the electric field reach its maximum magnitude?

$x = 0$ (at the center)
$x = \frac{R}{2}$
$x = \frac{R}{\sqrt{2}}$ (correct answer)
$x = R$
$x = R\sqrt{2}$

Explanation: When analyzing electric fields from symmetric charge distributions, you need to consider both the magnitude of the field contributions and their directions. For a charged ring, symmetry ensures the field points purely along the axis, but finding the maximum requires calculus. The electric field along the axis of a uniformly charged ring is given by

E = \frac{kQx}{(x^2 + R^2)^{3/2}}

, where

x

is the distance from the center along the axis. To find the maximum, you take the derivative and set it equal to zero:

\frac{dE}{dx} = \frac{kQ}{(x^2 + R^2)^{3/2}} - \frac{3kQx^2}{(x^2 + R^2)^{5/2}} = 0

. Solving this yields

x^2 = \frac{R^2}{2}

, so

x = \frac{R}{\sqrt{2}}

. Option A (

x = 0

) gives zero field because all charge elements are equidistant from the center, but their field vectors cancel due to symmetry. Option B (

x = \frac{R}{2}

) is too close to the ring—you haven't reached the optimal balance between distance and field strength. Option D (

x = R

) is too far; while the geometry is favorable, the

1/r^2

distance dependence weakens the field too much. The correct answer is C:

x = \frac{R}{\sqrt{2}}

represents the sweet spot where the axial distance is large enough to break the symmetry cancellation but not so large that the field becomes negligibly weak. Remember: Maximum field problems often involve taking derivatives of field expressions. The answer typically involves the characteristic length scale (here,

R

) modified by simple factors like

\sqrt{2}

Question 11

A cube of edge length $L$ is placed in a non-uniform electric field given by $\vec{E} = E_0\frac{x}{L}\hat{i}$ where $E_0$ is a constant. If one corner of the cube is at the origin, what is the electric flux through the face at $x = L$ ?

$E_0 L^2$ (correct answer)
$\frac{E_0 L^2}{2}$
$\frac{E_0 L^2}{4}$
$2E_0 L^2$
$0$

Explanation: When you encounter electric flux problems with non-uniform fields, you need to calculate the flux by integrating the electric field over the surface area. Electric flux is defined as

\Phi = \int \vec{E} \cdot d\vec{A}

, where

d\vec{A}

is the area element with direction normal to the surface. For the face at

x = L

, the area element points in the

+\hat{i}

direction, so

d\vec{A} = dx\,dy\,\hat{i}

where

dx

and

dy

are infinitesimal elements. Wait - that's wrong. Since we're looking at the face at

x = L

, this is a fixed surface, so

d\vec{A} = dy\,dz\,\hat{i}

. At the face where

x = L

, the electric field is

\vec{E} = E_0\frac{L}{L}\hat{i} = E_0\hat{i}

. Notice that the field is uniform across this entire face since

x

is constant at

L

. The flux calculation becomes:

\Phi = \int_0^L \int_0^L E_0\hat{i} \cdot dy\,dz\,\hat{i} = E_0 \int_0^L \int_0^L dy\,dz = E_0 L^2

This confirms answer A is correct. B (

\frac{E_0 L^2}{2}

) would result from incorrectly averaging the field over the range from 0 to

L

. C (

\frac{E_0 L^2}{4}

) might come from mistakenly using the average field value and then making an additional error with the area. D (

2E_0 L^2

) could result from double-counting or using the wrong area. Key strategy: In non-uniform field problems, always substitute the field value at the specific location of the surface you're analyzing. Don't get confused by the field's variation in regions you're not examining.

Question 12

An electric dipole consists of charges $+q$ and $-q$ separated by distance $d$ . At a point on the perpendicular bisector of the dipole, at distance $r >> d$ from the center, the electric field magnitude is approximately:

$\frac{kqd}{r^3}$ (correct answer)
$\frac{kq}{r^2}$
$\frac{2kq}{r^2}$
$\frac{kqd^2}{r^4}$
$0$

Explanation: When analyzing electric dipole problems, you need to consider both the magnitude and direction of the electric fields from each charge, especially when they're at comparable distances from your observation point. For a point on the perpendicular bisector at distance

r >> d

, each charge is approximately distance

r

away (since

r

is much larger than

d

). Each charge creates an electric field of magnitude

\frac{kq}{r^2}

. However, these fields don't simply add—you must account for their directions. The fields from the positive and negative charges point in different directions. When you break them into components, the components parallel to the dipole axis cancel out due to symmetry. Only the perpendicular components (pointing away from the dipole center) add constructively. Each field makes a small angle with the perpendicular bisector. For

r >> d

, this angle is approximately

\frac{d}{2r}

, so the perpendicular component of each field is

\frac{kq}{r^2} \cdot \frac{d}{2r} = \frac{kqd}{2r^3}

. Since both charges contribute equally, the total field is

\frac{kqd}{r^3}

. Choice B (

\frac{kq}{r^2}

) ignores the geometric cancellation—this would be the field from a single charge. Choice C (

\frac{2kq}{r^2}

) incorrectly adds the field magnitudes without considering direction. Choice D (

\frac{kqd^2}{r^4}

) has the wrong power of

d

and represents a higher-order correction term. Remember: dipole field problems always involve vector addition and geometric considerations. The key is recognizing when components cancel versus when they add constructively.

Question 13

A long cylindrical conductor of radius $a$ carries charge with linear density $\lambda$ . What is the electric field at distance $r > a$ from the axis?

$\frac{\lambda}{2\pi\epsilon_0 r^2}$
$\frac{\lambda}{2\pi\epsilon_0 r}$ (correct answer)
$\frac{\lambda}{4\pi\epsilon_0 r}$
$\frac{\lambda}{\pi\epsilon_0 r}$
$\frac{\lambda}{4\pi\epsilon_0 r^2}$

Explanation: When you encounter problems involving charged cylindrical conductors, this is a classic application of Gauss's law. The key insight is recognizing the cylindrical symmetry and choosing an appropriate Gaussian surface. For a long cylindrical conductor with linear charge density

\lambda

, you should construct a cylindrical Gaussian surface of radius

r > a

and length

L

, coaxial with the conductor. By symmetry, the electric field points radially outward and has constant magnitude at distance

r

. Applying Gauss's law:

\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}

. The left side equals

E \cdot 2\pi r L

(only the curved surface contributes since

\vec{E}

is perpendicular to the end caps). The enclosed charge is

Q_{enc} = \lambda L

. Therefore:

E \cdot 2\pi r L = \frac{\lambda L}{\epsilon_0}

, which gives

E = \frac{\lambda}{2\pi\epsilon_0 r}

. Looking at the wrong answers: Choice A (

\frac{\lambda}{2\pi\epsilon_0 r^2}

) has an extra factor of

r

in the denominator, suggesting confusion with point charge formulas where field falls as

1/r^2

. Choice C (

\frac{\lambda}{4\pi\epsilon_0 r}

) uses

4\pi

instead of

2\pi

, mixing up the surface area of a sphere with that of a cylinder. Choice D (

\frac{\lambda}{\pi\epsilon_0 r}

) is missing the factor of 2, possibly from incorrectly calculating the cylindrical surface area. Remember: cylindrical symmetry problems always yield

1/r

dependence, not

1/r^2

. The

2\pi r

factor comes directly from the circumference of your Gaussian cylinder.

Question 14

Consider a cube of side length $a$ with a point charge $+q$ located at one corner. What is the electric flux through the cube?

$\frac{q}{8\epsilon_0}$ (correct answer)
$\frac{q}{4\epsilon_0}$
$\frac{q}{2\epsilon_0}$
$\frac{q}{\epsilon_0}$
$\frac{2q}{\epsilon_0}$

Explanation: When you encounter electric flux problems with charges near or on the boundaries of surfaces, think about Gauss's law and how symmetry can help you visualize the solution. The key insight here is that a point charge at a corner of a cube creates the same electric field pattern as if you extended that charge's influence through symmetry. Imagine eight identical cubes arranged in a 2×2×2 configuration, with the point charge

+q

at the center where all eight cubes meet. By Gauss's law, the total flux through all eight cubes would be

\frac{q}{\epsilon_0}

. Since the charge is equidistant from all eight cubes and the geometry is perfectly symmetric, the flux must be equally distributed among them. Therefore, each cube receives

\frac{1}{8}

of the total flux, giving us

\frac{q}{8\epsilon_0}

. Looking at the wrong answers: Choice B (

\frac{q}{4\epsilon_0}

) represents the flux if the charge were at the center of a face, where it would be shared among 4 cubes. Choice C (

\frac{q}{2\epsilon_0}

) would apply if the charge were at the center of an edge, shared between 2 cubes. Choice D (

\frac{q}{\epsilon_0}

) is the total flux that would pass through a closed surface if the charge were entirely inside. Study tip: For charges on boundaries, use the symmetry method—imagine copying the geometry until the charge becomes centrally located, then divide the total flux by the number of identical regions you created. Corner = 8 regions, edge center = 4 regions, face center = 2 regions.

Question 15

A uniformly charged solid sphere of radius $R$ and total charge $Q$ has charge density $\rho = \frac{3Q}{4\pi R^3}$ . What is the electric field at distance $r = \frac{R}{3}$ from the center?

$\frac{Q}{12\pi\epsilon_0 R^2}$
$\frac{Q}{36\pi\epsilon_0 R^2}$
$\frac{Q}{108\pi\epsilon_0 R^2}$ (correct answer)
$\frac{Q}{4\pi\epsilon_0 R^2}$
$\frac{Q}{9\pi\epsilon_0 R^2}$

Explanation: When dealing with electric fields inside uniformly charged spheres, you need to apply Gauss's law with a key insight: only the charge enclosed within your Gaussian surface contributes to the field at that point. For a point at distance

r = \frac{R}{3}

from the center, construct a spherical Gaussian surface of radius

\frac{R}{3}

. The charge enclosed is only the portion within this smaller sphere, not the entire charge

Q

. The enclosed charge is:

Q_{enc} = \rho \cdot \frac{4}{3}\pi r^3 = \frac{3Q}{4\pi R^3} \cdot \frac{4}{3}\pi \left(\frac{R}{3}\right)^3 = \frac{3Q}{4\pi R^3} \cdot \frac{4\pi R^3}{81} = \frac{Q}{27}

Applying Gauss's law:

E \cdot 4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}

Therefore:

E = \frac{Q_{enc}}{4\pi \epsilon_0 r^2} = \frac{Q/27}{4\pi \epsilon_0 (R/3)^2} = \frac{Q/27}{4\pi \epsilon_0 R^2/9} = \frac{Q}{108\pi\epsilon_0 R^2}

This confirms answer C is correct. Answer A uses the wrong enclosed charge calculation, likely forgetting the cubic relationship in volume. Answer B makes an error in the geometric factors when computing the enclosed charge. Answer D incorrectly uses the total charge

Q

instead of just the enclosed portion, which would be the field outside the sphere. Remember: for electric fields inside charged spheres, always calculate what charge is actually enclosed by your Gaussian surface. The charge outside your surface doesn't contribute to the field at that point.

Question 16

Refer to the diagram. Four identical point charges $+q$ are arranged at the corners of a square of side length $a$ . What is the magnitude of the electric field at the center of the square?

$0$ (correct answer)
$\frac{kq}{a^2}$
$\frac{2kq}{a^2}$
$\frac{4kq}{a^2}$
$\frac{2\sqrt{2}kq}{a^2}$

Explanation: By symmetry, the electric field contributions from the four identical charges at the corners of the square cancel exactly at the center. Each charge is equidistant from the center, and the vector sum of four equal-magnitude vectors pointing radially outward from a symmetric arrangement is zero. Choices B-E represent various incorrect attempts to add the field magnitudes rather than the vector components.

Question 17

A hollow conducting sphere of radius $R$ has a small hole drilled through it. A point charge $+q$ is placed at the center of the sphere. Using Gauss's law, what can be concluded about the electric field just outside the hole?

The field is zero because the conductor shields the exterior region from the internal charge completely
The field has magnitude $\frac{kq}{R^2}$ pointing radially outward, same as if the hole weren't present (correct answer)
The field is undefined because Gauss's law cannot be applied when the conducting surface is not continuous
The field has magnitude greater than $\frac{kq}{R^2}$ because the hole concentrates field lines into a smaller area

Explanation: The point charge

+q

at the center induces charge

-q

on the inner surface of the conductor and

+q

on the outer surface. Even with a small hole, the charge distribution on the outer surface remains approximately uniform. Applying Gauss's law to a spherical surface just outside the conductor (enclosing total charge

+q

), we get

E \cdot 4\pi R^2 = q/\epsilon_0

, so

E = \frac{kq}{R^2}

. The hole is small enough that it doesn't significantly affect the overall field distribution.

Question 18

An infinite plane carries uniform surface charge density $\sigma > 0$ . A second infinite plane parallel to the first, separated by distance $d$ , carries uniform surface charge density $-2\sigma$ . What is the electric field magnitude in the region between the two planes?

$\frac{\sigma}{2\epsilon_0}$ pointing from the positive plane toward the negative plane
$\frac{3\sigma}{2\epsilon_0}$ pointing from the positive plane toward the negative plane (correct answer)
$\frac{\sigma}{\epsilon_0}$ pointing from the positive plane toward the negative plane
$\frac{\sigma}{2\epsilon_0}$ pointing from the negative plane toward the positive plane

Explanation: Each infinite plane creates field

E = \frac{|\sigma|}{2\epsilon_0}

pointing away from positive charge and toward negative charge. The

+\sigma

plane creates field

\frac{\sigma}{2\epsilon_0}

pointing away from it. The

-2\sigma

plane creates field

\frac{2\sigma}{2\epsilon_0} = \frac{\sigma}{\epsilon_0}

pointing toward it. Between the planes, both fields point in the same direction (from positive toward negative plane), so they add:

E_{total} = \frac{\sigma}{2\epsilon_0} + \frac{\sigma}{\epsilon_0} = \frac{3\sigma}{2\epsilon_0}

Question 19

Two identical conducting spheres of radius $R$ are initially uncharged and separated by a large distance. Sphere A is given charge $+Q$ , then brought into contact with sphere B, then separated again. Finally, a point charge $+q$ is placed at distance $d > R$ from the center of sphere B. What is the magnitude of the electric field at the center of sphere A due to all charges?

$\frac{kq}{(2d)^2}$ because sphere B acts like a point charge $+Q/2$ and only the external charge $+q$ contributes to field inside conductor A
$\frac{k(Q/2 + q)}{d^2}$ because the charges on sphere B and the external point charge combine to create the field at sphere A
$\frac{kQ}{4d^2} + \frac{kq}{d^2}$ because both sphere B and the point charge contribute to the external field at sphere A's location
Zero, because the field inside any conductor must always be zero regardless of external charge configurations (correct answer)

Explanation: When you encounter problems about electric fields and conductors, remember that conductors have a fundamental property: the electric field inside a conductor in electrostatic equilibrium is always zero, regardless of external charges or internal charge distribution. Let's trace through this problem. Initially, both spheres are uncharged. Sphere A gets charge

+Q

, then contacts sphere B. Since the spheres are identical, they share the charge equally, leaving each with

+Q/2

. When sphere B is later placed near point charge

+q

, the charges on sphere B will redistribute on its surface, but this doesn't change the fundamental conductor property. The key insight is that sphere A remains a conductor in electrostatic equilibrium throughout. The electric field at any point inside a conductor—including its center—must be zero. This is true regardless of what's happening outside: whether there's sphere B with charge

+Q/2

, point charge

+q

, or any other external configuration. Choice A incorrectly assumes only external charges matter while ignoring that we're inside a conductor. Choice B attempts to combine external charges as if we're in free space. Choice C calculates the field as if sphere A weren't a conductor at all. All these approaches miss the fundamental conductor property. The electric field inside conductor A is zero because free charges in the conductor arrange themselves on the surface to exactly cancel any internal field, no matter what external charges exist. Study tip: Whenever you see "electric field inside a conductor," immediately think "zero"—this is one of the most important and tested properties in electrostatics.

Question 20

A solid non-conducting sphere of radius $R$ has charge density $\rho(r) = \rho_0 \frac{r}{R}$ where $\rho_0$ is a constant. Using Gauss's law, what is the electric field magnitude at distance $r = R/2$ from the center?

$E = \frac{\rho_0 R}{6\epsilon_0}$ because the field scales directly with the average charge density in the enclosed volume
$E = \frac{\rho_0 R}{12\epsilon_0}$ because the linear density variation concentrates more charge in the outer regions
$E = \frac{\rho_0 R}{48\epsilon_0}$ because only the innermost quarter of the sphere contributes to the field at $r = R/2$
$E = \frac{\rho_0 R}{24\epsilon_0}$ because the enclosed charge varies as the cube of the radius for this density profile (correct answer)

Explanation: When applying Gauss's law to find electric fields with non-uniform charge distributions, you must carefully calculate the enclosed charge within your Gaussian surface, accounting for how the charge density varies with position. To find the electric field at

r = R/2

, construct a spherical Gaussian surface of radius

R/2

. The enclosed charge requires integrating the charge density over this volume:

Q_{enc} = \int_0^{R/2} \rho(r') \cdot 4\pi r'^2 dr' = \int_0^{R/2} \rho_0 \frac{r'}{R} \cdot 4\pi r'^2 dr'

Q_{enc} = \frac{4\pi\rho_0}{R} \int_0^{R/2} r'^3 dr' = \frac{4\pi\rho_0}{R} \cdot \frac{1}{4}\left(\frac{R}{2}\right)^4 = \frac{\pi\rho_0 R^3}{64}

Applying Gauss's law:

E \cdot 4\pi\left(\frac{R}{2}\right)^2 = \frac{Q_{enc}}{\epsilon_0}

Solving:

E = \frac{\pi\rho_0 R^3}{64\epsilon_0 \pi R^2} = \frac{\rho_0 R}{24\epsilon_0}

Choice A incorrectly assumes uniform charge density over the enclosed volume. Choice B contains an error in the integration or fails to properly account for the

r^3

dependence when integrating

\rho(r) \cdot r^2

. Choice C makes a fundamental error about which charge contributes to the field—all enclosed charge matters, not just a fraction. Choice D correctly recognizes that because

\rho(r) \propto r

, when integrated over spherical shells (

\propto r^2

), the enclosed charge depends on

r^4

, making the final field proportional to

r^2

inside the sphere. Study tip: For non-uniform charge distributions, always set up the integral for enclosed charge carefully, paying attention to how the charge density and volume element combine.