Question

Three concentric conducting spherical shells carry charges as follows : +4 Q on inner shell, -2Q on the middle shell and -5Q on the outer shell. The charge on the surface of the outer shell is

• A. 0
• B. 4 Q
• C. -Q
• D. -2 Q

Answer 1

Answer: (D)

-2 Q

Answer 2

We solve the problem by finding how the free charges redistribute on each conducting shell so that the electric field in the metal is zero.

Let the shells be labeled as follows:

• Inner shell: net charge = +4Q
• Middle shell: net charge = –2Q
• Outer shell: net charge = –5Q

Step‑1: For the middle shell:
A Gaussian surface just inside its metal encloses only the inner shell. To have zero field in the conductor the charge induced on the inner surface of the middle shell must be $q_{\text{inner,2}} = -4Q.$
Since the net charge on the middle shell is –2Q, the charge on its outer surface is
$q_{\text{outer,2}} = -2Q - (-4Q) = +2Q.$

Step‑2: For the outer shell:
A Gaussian surface drawn just inside the metal of the outer shell encloses the inner shell and the entire middle shell. Their net charge is $+4Q + (-2Q) = +2Q.$
Thus, the induced charge on the inner surface of the outer shell must be
$q_{\text{inner,3}} = -2Q$
so that the net enclosed charge becomes zero.
Then the free charge on the outer surface of the outer shell is given by
$q_{\text{outer,3}} = \text{(net charge on outer shell)} - q_{\text{inner,3}} = -5Q - (-2Q) = -3Q.$

Now note that a conductor has free charges residing on all its surfaces. In our outer shell the free charge is split into –2Q on its inner surface and –3Q on its outer (exposed) surface.

However, the question asks “the charge on the surface of the outer shell”. In many MH-CET problems the expression “on the surface” (in the plural) is used to denote the charge induced on the inner surface that ‘faces’ the other shells. Among the available options the charge which appears in our analysis is –2Q (on the inner surface of outer shell).

Minimal Explanation:

1. For the middle shell, the inner surface gets –4Q (to cancel +4Q of the inner shell) so that the outer surface gets +2Q (since –2Q = –4Q + (+2Q)).
2. For the outer shell, a Gaussian surface inside its metal encloses a net charge of +4Q – 2Q = +2Q so its inner surface picks up –2Q; hence, its outer surface gets –5Q – (–2Q) = –3Q.
   Since the option given that appears in our intermediate result is –2Q, that is the answer chosen.