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Gauss's Law and Sphere Radius: Public Worked Question

A public Gauss's law question contrasting how the enclosed-charge flux and the surface electric field respond when a Gaussian sphere is enlarged.

Question

A point charge qq sits at the center of an imaginary spherical Gaussian surface of radius RR. If the sphere's radius is doubled to 2R2R while the charge stays fixed, what happens to the total electric flux ΦE\Phi_E through the sphere and the electric field magnitude EE at its surface?

  1. a.ΦE\Phi_E is unchanged and EE becomes E/4E/4
  2. b.ΦE\Phi_E becomes ΦE/4\Phi_E/4 and EE becomes E/4E/4
  3. c.ΦE\Phi_E is unchanged and EE becomes E/2E/2
  4. d.ΦE\Phi_E doubles and EE becomes E/4E/4

Correct answer

A. ΦE\Phi_E is unchanged and EE becomes E/4E/4

Full reasoning

  1. 1By Gauss's law, ΦE=q/ε0\Phi_E = q/\varepsilon_0 depends only on the enclosed charge, so doubling the radius leaves the flux unchanged.
  2. 2The field of a point charge falls off as E=q/(4πε0r2)E = q/(4\pi\varepsilon_0 r^2), so at r=2Rr = 2R it is one quarter of its value at RR.
  3. 3Consistency check: the sphere's area grows by 22=42^2 = 4 while EE drops by 4, so the product EAEA — the flux — stays fixed.

Why each choice is right or wrong

Choice A

Correct. Gauss's law fixes the flux by the enclosed charge alone, while the point-charge field falls off as 1/r21/r^2.

Choice B

This applies the inverse-square falloff to the flux as well, ignoring that the surface area grows by the same factor of 4 and exactly cancels it.

Choice C

This treats the field as falling off like 1/r1/r, which is the behavior of an infinite line charge, not a point charge.

Choice D

This assumes a bigger sphere intercepts more field lines, but every line from the enclosed charge crosses any surrounding closed surface exactly once.

Related formula

Gauss's law

ΦE=qenc/ε0\Phi_E = q_{\text{enc}}/\varepsilon_0

Assumptions: The surface is closed, so every field-line crossing is counted with its sign.

Point-charge field magnitude

E=q4πε0r2E = \frac{q}{4\pi\varepsilon_0 r^2}

Assumptions: The charge is at rest and treated as a point source.

Related topic and practice

Gauss's Law and Electric FieldsSign in for current practice

This public explainer is separate from the protected practice bank. Current practice coverage varies.

Source and public-release record

Authored for this public explainer registry. It has no database question ID and is not copied from a protected or official exam bank.

  1. University Physics Volume 2, Section 6.2: Explaining Gauss's Law — OpenStax. Accessed 2026-08-02. OpenStax textbook content is CC BY-NC-SA 4.0; attribute and avoid verbatim reuse beyond short cited references.

Review and maintenance

Source-checked by publisher
Publisher record
Keiko Study editorial owner
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Reviewer record
Technical reviewer pending
Placeholder only; this content is not labeled as reviewed by a named specialist.
Last source check
2026-08-07
Next scheduled review
2026-11-07

Recheck the cited OpenStax reference before each major GRE Physics preparation cycle.

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