📝 Gauss's law for inverse square fields (16 MCQs)

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

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What is Gauss's law for inverse square fields?

Gauss's law for inverse square fields:
For $\mathbf{F} = k \frac{\mathbf{r}}{r^3}$ , $\nabla \cdot \mathbf{F} = 0$ for $r>0$ , but flux through any closed surface enclosing the origin equals $4\pi k$ .

Example:
Electric field of a point charge $\mathbf{E} = \frac{q}{4\pi \epsilon_0} \frac{\mathbf{r}}{r^3}$ has flux $q/\epsilon_0$ through any closed surface.

Reason:
This is Gauss's law, showing inverse square fields have non-zero flux only at sources, fundamental to electrostatics.

Easy

Medium

Hard

📝 All Gauss's law for inverse square fields MCQs

Q1. A vector field has the form $\mathbf{F}=k\frac{\hat{\mathbf{r}}}{r^2}$ away from the origin. What feature most directly makes the total outward flux through any sphere centered at the origin independent of the sphere's radius?

A.The field magnitude is constant on every sphere

B.The spherical area increases as

r^2

, compensating for the

1/r^2

decrease ✅

C.The field becomes zero at large distances

D.The field direction changes randomly with radius

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The field magnitude decreases as $1/r^2$ , while the surface area of a sphere increases as $4\pi r^2$ . Their product therefore remains constant. This geometric cancellation is the key reason the total outward flux through centered spheres does not depend on their radius.

Q2. A student claims that because an inverse-square field becomes weaker with distance, the outward flux through a larger spherical surface must also decrease. Which response best identifies the flaw?

A.Flux depends only on field magnitude

B.The larger surface has proportionally more area, exactly compensating for the weaker field ✅

C.Flux is independent of both field and area

D.The field direction becomes stronger at larger distances

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: The student's reasoning considers only the local field strength and ignores the growing area of the surface. For a sphere, area scales as $r^2$ , while the inverse-square field scales as $1/r^2$ . These factors cancel, leaving the total flux unchanged.

Q3. Two concentric spherical surfaces have radii $R$ and $3R$ and enclose the same source. If the field is purely radial and follows an inverse-square dependence, how do their total outward fluxes compare?

A.The larger sphere has nine times the flux

B.The larger sphere has one-third the flux

C.Both spheres have the same total flux ✅

D.The smaller sphere has nine times the flux

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: At radius $R$ , the field is proportional to $1/R^2$ , while the area is proportional to $R^2$ . At $3R$ , the field becomes $1/9$ as large, but the area becomes nine times larger. Multiplying field by area therefore gives the same total flux.

Q4. A point source produces an outward inverse-square field. A closed spherical surface is expanded so that its radius doubles, while the source remains at the center. What happens to the field magnitude and total flux, respectively?

A.Field doubles; flux doubles

B.Field becomes four times smaller; flux remains unchanged ✅

C.Field becomes two times smaller; flux becomes four times smaller

D.Field remains unchanged; flux becomes four times larger

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Doubling the radius changes the inverse-square field by a factor of $1/2^2=1/4$ . However, the spherical area increases by $2^2=4$ . Thus the local field becomes four times weaker, but the fourfold increase in area compensates exactly, keeping the total flux unchanged.

Q5. A source is placed at the center of a spherical surface. An experimenter measures field magnitude at several radii and obtains $E\propto 1/r^2$ . Which additional observation would most strongly support the conclusion that the total outward flux is constant?

A.The field direction is radial and the measured flux product

E(4\pi r^2)

stays constant ✅

B.The field direction changes with radius while

E r

stays constant

C.The field magnitude is constant while surface area increases

D.The product

Er^2

decreases steadily with radius

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: For a centered spherical surface, the flux is obtained from the field magnitude multiplied by the surface area. If $E\propto1/r^2$ , then $E(4\pi r^2)$ should remain constant. A radial field also ensures that the field is aligned with the outward surface direction everywhere.

Q6. A point source produces an inverse-square field. A small spherical detector is moved from distance $r$ to distance $2r$ . The detector's area is kept fixed rather than expanded with distance. What happens to the flux through the detector, assuming its orientation remains perpendicular to the field?

A.It remains unchanged

B.It becomes four times larger

C.It becomes one-fourth as large ✅

D.It becomes one-half as large

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: The important distinction is between flux through a fixed small detector and total flux through a complete sphere. At $2r$ , the inverse-square field is one-fourth of its value at $r$ . Since the detector area is unchanged, its flux also becomes one-fourth as large.

Q7. A researcher surrounds a source with a closed surface that is irregular rather than spherical. The source remains completely inside the surface. Which conclusion is most justified if the field follows an inverse-square radial behavior from the source?

A.The total outward flux must be zero because the surface is irregular

B.The total outward flux can remain the same as for a sphere because flux depends on how the field spreads over the closed surface ✅

C.The flux must depend only on the largest radius of the surface

D.The flux is necessarily larger because the irregular surface has greater area

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Spherical symmetry makes the calculation simple, but the physical conservation of an inverse-square field is not restricted to spherical surfaces. A distorted closed surface intercepts different field strengths over different areas. Where the surface is farther away, the field is weaker, while the intercepted area can compensate correspondingly.

Q8. A student calculates the flux through a sphere by multiplying the field at the sphere's center by the sphere's surface area. The source is located at the center, so the field at that point is undefined. What is the better strategy?

A.Use the field at a point just outside the center

B.Evaluate the field on the spherical surface, where its magnitude is constant, and integrate over the surface ✅

C.Replace the source by a uniform field

D.Ignore the singular point because it has zero area and use zero field everywhere

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: The field at the source location is not the appropriate value for calculating surface flux. Instead, the field must be evaluated on the actual surface. For a centered sphere, every surface point is at the same radius, so the inverse-square field has constant magnitude there and is radial.

Q9. A graph of field magnitude $E$ versus radius $r$ is plotted on ordinary linear axes. The curve falls rapidly near the source and then flattens. Which interpretation is most consistent with an inverse-square field?

A.The field varies approximately as

r^2

B.The field varies approximately as

1/r

C.The field varies approximately as

1/r^2

✅

D.The field is constant after a short distance

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: An inverse-square dependence decreases very rapidly near small radii and becomes progressively flatter as radius increases. Thus a graph of $E$ against $r$ has a steep decline near the source followed by a gradual approach toward zero. This shape is characteristic of $E\propto1/r^2$ .

Q10. A graph shows $E r^2$ on the vertical axis and $r$ on the horizontal axis for measurements around a centered source. The graph is approximately horizontal. What physical conclusion is most reasonable?

A.The field is proportional to

r^2

B.The measurements are consistent with an inverse-square field ✅

C.The source strength increases with distance

D.The field has no radial dependence

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If $Er^2$ is approximately constant, then $E$ must approximately vary as $1/r^2$ . This is a direct graphical test of inverse-square behavior. Experimental deviations from a perfectly horizontal graph would indicate measurement uncertainty or departures from the idealized field model.

Q11. Two students use different methods to determine the total flux from a point source. Student A integrates the field over a sphere centered on the source. Student B surrounds the source with a complicated closed surface and argues from conservation of field spreading. Under ideal inverse-square behavior, what should happen?

A.Only Student A can obtain the correct total flux

B.Only Student B can obtain the correct total flux

C.Both can obtain the same total flux if their closed surfaces enclose the same source ✅

D.Their results must differ because their surfaces have different areas

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: A centered sphere provides an especially convenient calculation because the field magnitude is constant over its surface. However, the total outward flux associated with an inverse-square source is determined by the enclosed source, not by the arbitrary shape of a closed enclosing surface. Both approaches can therefore agree.

Q12. An inverse-square field is generated by a source at the origin. A closed surface is moved outward until it encloses an additional source of opposite sign. How should the net outward flux change compared with the original surface?

A.It must remain unchanged because all inverse-square fields conserve flux

B.It changes according to the combined signed contributions of the enclosed sources ✅

C.It becomes zero regardless of source strengths

D.It depends only on the distance between the two sources

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Each source contributes to the total outward flux according to its signed strength. When an additional source of opposite sign becomes enclosed, its contribution is added algebraically and therefore reduces the original net flux. The final value depends on the combined strengths, not simply on the number of sources.

Q13. A student argues: 'If the field at radius $2R$ is one-fourth the field at $R$ , then the flux through the larger sphere must also be one-fourth.' Which missing step invalidates this conclusion?

A.The larger sphere has a surface area four times as large ✅

B.The smaller sphere has a field direction opposite to the larger sphere

C.Flux is unrelated to field magnitude

D.The source disappears when the radius increases

💡 Difficulty: medium | ✅ Correct: A

📖 Explanation: The field comparison alone is correct, but flux also depends on the surface area. Increasing the radius from $R$ to $2R$ makes the field four times weaker while making the spherical area four times larger. Therefore the two effects cancel for the total flux.

Q14. A numerical model predicts $E(r)=K/r^2$ for $r>0$ . A second model predicts $E(r)=K/(r^2+R^2)$ . At very large $r$ , both models appear similar. Which test would best distinguish their predicted total flux through increasingly large centered spheres?

A.Check whether

E(4\pi r^2)

approaches a constant ✅

B.Check only the field direction at

r=R

C.Compare the field at one small radius only

D.Measure the sphere's diameter without measuring the field

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: For the exact inverse-square model, multiplying field by spherical area gives $E(4\pi r^2)=4\pi K$ , a constant at every radius. The second model gives $4\pi K r^2/(r^2+R^2)$ , which approaches a constant only asymptotically. Testing several radii can therefore distinguish the models.

Q15. A point source creates an outward field proportional to $1/r^2$ . Suppose the source strength is increased by a factor of three while the observation sphere remains centered on the source. Which combined change is expected?

A.Field decreases by three and flux decreases by three

B.Field increases by three and total flux increases by three ✅

C.Field increases by nine and total flux remains unchanged

D.Field remains unchanged and total flux triples

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: The field magnitude is directly proportional to the source strength, so tripling the source strength triples the field at every fixed radius. Because the spherical area has not changed, the total flux also triples. The inverse-square dependence controls radial variation, while source strength controls overall magnitude.

Q16. Consider a hypothetical radial field $E(r)=K/r^n$ . For which value of $n$ does the total flux through a centered sphere remain independent of radius, assuming the spherical area scales as $4\pi r^2$ ?

n=0

n=1

n=2

✅

n=3

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: The total flux scales as field magnitude times area, giving $\Phi\propto r^{-n}r^2=r^{2-n}$ . For the flux to be independent of radius, the exponent must be zero, so $2-n=0$ . Therefore $n=2$ , identifying the inverse-square dependence as the special scaling that preserves constant total flux.

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