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Home/Practice questions/Physics
PhysicsElectric Charges and Fields

Given below are two statements: one is labelled as Assertion (A) and…

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): The electric field is zero at every point inside a uniformly charged thin spherical shell.

Reason (R): A spherical Gaussian surface drawn through any point inside the shell, with its centre at the centre of the shell, encloses no charge.

In the light of the above statements, choose the most appropriate answer from the options given below:

Answer & explanation+

Correct answer: option 1

The correct answer is Option (1) → Both (A) and (R) are correct and (R) is the correct explanation of (A).

Assertion (A) The electric field is zero at every point inside a uniformly charged thin spherical shell. — True. This holds at every point inside the shell.

Reason (R) A spherical Gaussian surface drawn through any point inside the shell, with its centre at the centre of the shell, encloses no charge. — True. All the charge of a thin shell sits on the shell itself, outside such a Gaussian sphere.

Why Reason (R) explains Assertion (A)?

  • By spherical symmetry, the field on the Gaussian sphere of radius \( r \) has the same magnitude everywhere and is radial. The flux through it is \( E\times4\pi r^{2} \).

  • The enclosed charge is zero, so Gauss’s law gives \( E\times4\pi r^{2}=0 \). Hence \( E=0 \) at every point inside.

NCERT (2026-27):
Field due to a uniformly charged thin spherical shell – “The flux through the Gaussian surface, calculated as before, is E × 4πr2. However, in this case, the Gaussian surface encloses no charge. Gauss’s law then gives E × 4πr2 = 0 i.e., E = 0 (r < R ) … that is, the field due to a uniformly charged thin shell is zero at all points inside the shell …”

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