📝 Gauss's law in electrostatics (15 MCQs)

📖 From Calculus • 16. Topics in vector Calculus • 15 questions available

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What is Gauss's law in electrostatics?

Gauss's law in electrostatics:
$\oint_S \mathbf{E} \cdot d\mathbf{S} = \frac{Q_{\text{enc}}}{\epsilon_0}$ states the total electric flux through a closed surface equals the enclosed charge divided by $\epsilon_0$ .

Example:
For a point charge $q$ at origin, flux through a sphere radius $a$ is $q/\epsilon_0$ , independent of sphere radius.

Reason:
Gauss's law is one of Maxwell's equations, providing a powerful method to compute electric fields for symmetric charge distributions.

Easy

Medium

Hard

📝 All Gauss's law in electrostatics MCQs

Q1. A closed surface encloses a net electric charge $Q_{\mathrm{enc}}$ . Which quantity determines the total electric flux through the surface?

A.The shape and volume of the surface

B.The net charge enclosed by the surface ✅

C.The charges located outside the surface only

D.The electric field at the geometric center

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Gauss's law states that the net electric flux through any closed surface depends only on the net charge enclosed, according to $\Phi_E=Q_{\mathrm{enc}}/\varepsilon_0$ . The surface shape, volume, and external charges can affect local field values but do not change the net flux.

Q2. A spherical Gaussian surface is enlarged while remaining centered on the same isolated point charge. What happens to the total electric flux through the surface?

A.It increases because the surface area increases

B.It decreases because the electric field becomes weaker

C.It remains unchanged because the enclosed charge is unchanged ✅

D.It becomes zero because the field is nonuniform on the larger surface

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: The electric field magnitude decreases as the sphere expands, but the area increases in such a way that their product remains consistent with the enclosed charge. Since the enclosed charge is unchanged, Gauss's law requires the total flux to remain $Q/\varepsilon_0$ .

Q3. A closed Gaussian surface is moved through a region containing several charges, but no charge crosses the surface boundary during the motion. Which conclusion is necessarily correct?

A.The total electric flux remains unchanged ✅

B.The electric field must remain unchanged everywhere

C.The flux must become zero

D.The surface must remain spherical

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The total flux is fixed by the net enclosed charge rather than by the surface's location or shape. If no charge crosses the boundary, $Q_{\mathrm{enc}}$ remains constant, so the net flux remains constant even though the electric field distribution over the surface may change.

Q4. A student argues: "If the electric field is zero at every point on a closed surface, then the surface cannot enclose any charge." Which assessment is most accurate?

A.The reasoning is always correct because zero field means zero charge everywhere

B.The reasoning is incomplete because the field could be zero on the boundary only under special charge configurations ✅

C.The reasoning is wrong because electric flux never depends on charge

D.The reasoning is wrong because a closed surface always has positive flux

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If the field is exactly zero everywhere on the closed surface, the flux through that surface is zero, so the net enclosed charge must be zero. However, that does not mean there can be no individual positive and negative charges inside. Their net charge could cancel, making the student's statement too strong.

Q5. A closed surface encloses charges $+3q$ , $-q$ , and $-2q$ . A separate charge $+5q$ is located outside the surface. What is the net electric flux through the surface?

5q/\varepsilon_0

B.0 ✅

3q/\varepsilon_0

7q/\varepsilon_0

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The enclosed charges add to $+3q-q-2q=0$ . Therefore $Q_{\mathrm{enc}}=0$ , and Gauss's law gives $\Phi_E=Q_{\mathrm{enc}}/\varepsilon_0=0$ . The external $+5q$ can alter the electric field at points on the surface, but it contributes no net enclosed charge.

Q6. A spherical Gaussian surface encloses a charge $Q$ . The surface is distorted into an irregular closed shape without crossing the charge. Which statement best describes the new total flux?

A.It must increase because the irregular surface has more area

B.It must decrease because the surface is no longer spherical

C.It remains

Q/\varepsilon_0

, although the field distribution may become complicated ✅

D.It becomes dependent on the maximum distance from the charge

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Gauss's law does not require the Gaussian surface to be spherical. As long as the surface remains closed and continues to enclose the same net charge $Q$ , the total flux is $Q/\varepsilon_0$ . What changes is generally the local flux density and field orientation over the surface.

Q7. A point charge $Q$ is enclosed by a cube whose center is at the charge location. The cube is then rotated about its center without changing its size or moving the charge. What happens to the total flux?

A.It changes because the faces change orientation

B.It doubles because the charge is closer to some edges

C.It remains

Q/\varepsilon_0

because the enclosed charge is unchanged ✅

D.It becomes zero because opposite faces cancel

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Rotation changes the orientation of each face and therefore changes individual flux contributions in general. However, the total flux through the complete closed surface is determined only by the enclosed charge. Since the charge remains inside the cube, the total flux remains $Q/\varepsilon_0$ .

Q8. A closed surface initially encloses net charge $2Q$ . A charge $Q$ is moved from inside the surface to outside without crossing the surface itself; instead, imagine the charge is removed from the interior and placed outside. What is the new total flux?

3Q/\varepsilon_0

2Q/\varepsilon_0

Q/\varepsilon_0

✅

D.0

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Initially the enclosed charge is $2Q$ . Removing one charge $Q$ from the interior changes the enclosed net charge to $Q$ , while placing that charge outside does not add to $Q_{\mathrm{enc}}$ . Therefore the new total flux is $Q/\varepsilon_0$ , regardless of the outside charge's field.

Q9. A student calculates the electric flux through each face of a cube caused by an external point charge and concludes that the total flux must equal the flux expected from that point charge. What is the key error?

A.External charges can contribute to local flux but cannot determine net flux through the closed surface ✅

B.Electric flux can only be calculated for spherical surfaces

C.External charges always produce zero electric field on closed surfaces

D.The cube must contain a charge before any electric flux can exist

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: An external charge can produce a substantial electric field on every face of a closed surface, so individual flux contributions need not vanish. However, when all faces are considered together, the external charge contributes zero net flux because it is not enclosed. Only the net enclosed charge determines total flux.

Q10. A graph shows the electric flux $\Phi_E$ through several closed surfaces plotted against enclosed net charge $Q_{\mathrm{enc}}$ . The data form a straight line passing through the origin. If the slope is $1/\varepsilon_0$ , what physical conclusion follows?

A.Flux is independent of charge

B.The graph contradicts Gauss's law

C.The measured relationship is consistent with

\Phi_E=Q_{\mathrm{enc}}/\varepsilon_0

✅

D.The slope represents the electric field magnitude

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The graph directly represents the linear relationship between total electric flux and enclosed charge. A slope of $1/\varepsilon_0$ gives $\Phi_E/Q_{\mathrm{enc}}=1/\varepsilon_0$ , which matches Gauss's law. The slope is not the electric field itself because electric field and total flux have different physical meanings and units.

Q11. Two closed surfaces $S_1$ and $S_2$ have very different shapes. $S_1$ encloses charges $+4Q$ and $-Q$ , while $S_2$ encloses only $+Q$ . Which comparison is correct?

S_1

has four times the flux of

S_2

S_1

has three times the flux of

S_2

✅

C.Both surfaces have the same flux

D.The flux through

S_2

is three times that through

S_1

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: For $S_1$ , the net enclosed charge is $4Q-Q=3Q$ , so its flux is $3Q/\varepsilon_0$ . For $S_2$ , the enclosed charge is $Q$ , giving $Q/\varepsilon_0$ . Thus the flux through $S_1$ is three times the flux through $S_2$ , independent of their different shapes.

Q12. A closed surface has a net outward flux of $8Q/\varepsilon_0$ . Later, the surface is enlarged while remaining closed and no charge crosses its boundary. Which statement best predicts the new net flux?

A.It becomes

16Q/\varepsilon_0

because the area doubles

B.It becomes

4Q/\varepsilon_0

because the field weakens

C.It remains

8Q/\varepsilon_0

because the enclosed charge has not changed ✅

D.It becomes zero because the surface is larger

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The original flux corresponds to an enclosed net charge of $8Q$ . Enlarging the surface without allowing charge to cross its boundary leaves that enclosed charge unchanged. Although the electric field may become weaker at many points and the area changes, the complete-surface flux remains $8Q/\varepsilon_0$ .

Q13. A charged object is surrounded by a closed Gaussian surface. Measurements show that the electric field is outward over some portions of the surface and inward over others. A student concludes that Gauss's law cannot be applied because the field changes direction. What is the best response?

A.The student is correct because Gauss's law requires a uniform field

B.The student is correct because inward flux is impossible in electrostatics

C.The conclusion is incorrect because Gauss's law sums signed flux over the entire closed surface ✅

D.The conclusion is incorrect only if the surface is spherical

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: Gauss's law does not require the electric field to have a uniform direction over the surface. Flux is a signed quantity involving the dot product of the field with the outward area vector. Regions with inward and outward contributions can partially cancel, leaving a net value determined by enclosed charge.

Q14. A closed surface contains several charges whose algebraic sum is zero. An external charge is then brought very close to the surface without crossing it. Which outcome is possible?

A.The net flux becomes nonzero because the external field becomes stronger

B.The net flux remains zero even though the local flux density can change substantially ✅

C.Both the net flux and every local flux contribution must remain unchanged

D.The net flux becomes infinite when the external charge approaches the surface

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The external charge can strongly distort the electric field on the surface, so local flux through different portions may change dramatically. Nevertheless, the external charge remains outside the closed surface, while the net enclosed charge remains zero. Therefore Gauss's law requires the total net flux to remain zero.

Q15. Consider a family of nested closed surfaces centered on the same point. The enclosed charge changes from $Q$ to $2Q$ to $4Q$ as successive charges become enclosed. Which qualitative graph of total flux versus enclosed charge must result?

A.A horizontal line

B.A straight line through the origin with constant positive slope ✅

C.A parabola opening upward

D.A straight line with decreasing slope

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Gauss's law gives $\Phi_E=Q_{\mathrm{enc}}/\varepsilon_0$ , so total flux is directly proportional to enclosed charge. Therefore a graph of flux against enclosed charge must be linear and pass through the origin, with constant slope $1/\varepsilon_0$ . Doubling the enclosed charge doubles the total flux.

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