Two charges $q_1$ and $q_2$ are placed at the centres of two spherical conducting shells of radius $r_1$ and $r_2$ respectively. The shells are arranged such that their centres are $d\ [>(r_1+r_2)]$ distance apart. The force on $q_2$ due to $q_1$ is : (A) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{d^2}$ (B) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{(d-r_1)^2}$ (C) Zero (D) $\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{[d-(r_1+r_2)]^2}$

Two charges q₁ and q₂ are placed at the centres of two spherical conducting shells of radius r₁ and r₂ respect

Question

q_1

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★

✓ Free question

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.
So the field inside the cavity — where $q_2$ actually sits — is due only to $q_2$ itself and the uniform $-q_2$ layer on the inner wall. A uniformly charged spherical shell produces zero field at its own centre. Excluding $q_2$ 's (undefined) self-force, the net field acting on $q_2$ from everything else is exactly zero.

\vec{E}_{\text{at }q_2,\ \text{due to everything outside shell 2}} = \vec{0}

Therefore the force on $q_2$ due to $q_1$ is zero:

\vec{F} = q_2\,\vec{E} = q_2 \times \vec{0} = \vec{0}.

Watch out

The common mistake — treating both shells as simple point charges and writing $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{d^2}$ — ignores that $q_2$ is shielded by its own shell. That formula would be correct only if $q_2$ were a bare charge sitting directly in the external field, with no conducting shell of its own around it.

Note

Note the asymmetry: shell 1 (and $q_1$ ) still influence the outer surface of shell 2 (redistributing its charge, producing a net force on shell 2 as a whole through the induced charges). What is exactly zero is the force on $q_2$ itself, because $q_2$ is shielded inside the cavity — this is the specific, deliberately-tested subtlety of the question.

✓Final answer

q_2

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

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.

So the field inside the cavity — where $q_2$ actually sits — is due only to $q_2$ itself and the uniform $-q_2$ layer on the inner wall. A uniformly charged spherical shell produces zero field at its own centre. Excluding $q_2$ 's (undefined) self-force, the net field acting on $q_2$ from everything else is exactly zero.

$\vec{E}_{\text{at }q_2,\ \text{due to everything outside shell 2}} = \vec{0}$

Therefore the force on $q_2$ due to $q_1$ is zero:

$\vec{F} = q_2\,\vec{E} = q_2 \times \vec{0} = \vec{0}.$
Watch out

The common mistake — treating both shells as simple point charges and writing $F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{d^2}$ — ignores that $q_2$ is shielded by its own shell. That formula would be correct only if $q_2$ were a bare charge sitting directly in the external field, with no conducting shell of its own around it.

Note

Note the asymmetry: shell 1 (and $q_1$ ) still influence the outer surface of shell 2 (redistributing its charge, producing a net force on shell 2 as a whole through the induced charges). What is exactly zero is the force on $q_2$ itself, because $q_2$ is shielded inside the cavity — this is the specific, deliberately-tested subtlety of the question.

✓Final answer
$q_2$

Why the field cannot reach q2q_2q2​

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Why the field cannot reach $q_2$

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