Write a short note on 'electrostatic shielding'.

II. Short Answer Questions · Q17

Q. Write a short note on 'electrostatic shielding'.

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Step 1. For a conductor with a hollow cavity carved out of it, the field inside the cavity is exactly zero as long as no charge is placed inside the cavity, regardless of how strong an external field is applied outside the conductor.

Step 2. This is because induced surface charges on the conductor cancel the field throughout the conducting material and, consequently, inside the enclosed cavity too. This complete blocking of external fields from a hollow conductor's interior …

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CBSE 2023Set 55/4/11 markMCQ
$q_1$
›Reveal solutionSolution
Because $q_2$ sits at the centre of its own conducting shell, that shell electrostatically shields $q_2$ from any external field — including the field of $q_1$ and its shell. The force on $q_2$ due to $q_1$ is $\boxed{\text{Zero}}$ , option (C).

The core concept here is electrostatic shielding: the field inside the cavity of a conductor depends only on the charge enclosed within that cavity, never on anything outside the conductor — however strong, however close. This is exactly the property that makes a Faraday cage work, and NCERT states it explicitly for the cavity-in-a-conductor case.

Important

It is tempting to treat each shell-plus-central-charge system as an effective point charge (true for the field outside an isolated shell) and then simply plug into Coulomb's law. That reasoning correctly describes the field outside shell 1 at large distances. But it silently assumes $q_2$ is sitting bare in that field — it is not. $q_2$ is enclosed within its own conducting shell (shell 2), and that changes everything about the force actually experienced by $q_2$ .

Why the field cannot reach $q_2$

Shell 2 is a conductor in electrostatic equilibrium, so the field is zero everywhere within the metal of shell 2 itself.

The charge on the inner surface of shell 2 is fixed by $q_2$ alone. Because $q_2$ sits exactly at the centre of the spherical cavity, the induced charge on the inner wall must be a uniform layer of total charge $-q_2$ — this is the only distribution consistent with (a) the field being zero in the conductor bulk, and (b) $q_2$ being at the geometric centre. This uniform layer is completely determined by $q_2$ 's position; it has nothing to do with $q_1$ or shell 1.

Whatever the external field from shell 1 does, it can only rearrange the outer surface charge of shell 2 (making it non-uniform, so that shell 2's outer surface, seen from far away, is no longer a perfect point-charge equivalent). This outer-surface rearrangement is exactly what "soaks up" the external field — it cannot communicate anything back through the conducting bulk to the inner surface or the cavity. …
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Q.Write a short note on 'electrostatic shielding'.

Concept understanding — Electrostatic Shielding

Why the field cannot reach $q_2$

Q.Write a short note on 'electrostatic shielding'.

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