Problem 27 - A Complete Resource Book in Physics for JEE Main

Question

A spherical conductor \(A\) of radius \(r\) is placed concentrically inside a conducting shell \(B\) of radius \(R(R>r)\). A charge \(Q\) is given to \(A\), and then \(A\) is joined to \(B\) by a metal wire. The charge flowing from \(A\) to \(B\) will be (A) \(Q\left(\frac{R}{R+r}\right)\) (B) \(Q\left(\frac{r}{R+r}\right)\) (C) \(Q\) (D) Zero

Step-by-Step Solution

Answer

The charge flowing to conductor B is: \(Q_{flow} = Q\left(\frac{R}{R + r}\right)\). Thus, the correct answer is (A).

Step 1: Understand the problem and initial conditions

We are given a system of two concentric spherical conductors, A and B. Conductor A has a radius r and is initially charged with charge Q. Conductor B has a larger radius of R, and is initially uncharged. The conductors are connected by a metal wire.

Step 2: Determine the electric potential before connection

Before connecting the metal wire, the electric potential of conductor A is given by \(V_A = \frac{Q}{4\pi \epsilon_0 r}\), where \(\epsilon_0\) is the permittivity of free space. Conductor B is initially uncharged, so its electric potential is zero.

Step 3: Connect the metal wire and find the redistribution of charge

When the metal wire is connected, the electric potential of both conductors becomes equal, as they are now acting as a single conductor. Let's denote the total charge on A after connection as Q1 and the total charge on B as Q2. According to the law of charge conservation, we have the equation: \[Q1 + Q2 = Q\]

Step 4: Determine the new electric potentials

Since the conductors are now connected, their electric potentials are equal (\(V_A = V_B\)). The electric potentials for conductors A and B are given by: \[V_A = \frac{Q1}{4\pi \epsilon_0 r}\] \[V_B = \frac{Q2}{4\pi \epsilon_0 R}\] We can now equate these expressions: \[\frac{Q1}{r} = \frac{Q2}{R}\]

Step 5: Solve for Q1 and Q2

Now, we have two equations: \[Q1 + Q2 = Q\] \[\frac{Q1}{r} = \frac{Q2}{R}\] Let's solve for Q2: \[Q2 = Q - Q1\] Substitute this into the second equation: \[\frac{Q1}{r} = \frac{Q - Q1}{R}\] Rearrange and solve for Q1: \[Q1(R + r) = Qr\] \[Q1 = \frac{Qr}{R + r}\] Now, substitute back into the equation for Q2: \[Q2 = Q - \frac{Qr}{R + r} = Q\left(\frac{R}{R + r}\right)\]

Step 6: Calculate the charge flowing to conductor B

Finally, the charge flowing to conductor B is given by the difference between the initial charge on A and the final charge on A: \[Q_{flow} = Q - Q1 = Q\left(1 - \frac{r}{R + r}\right) = Q\left(\frac{R}{R + r}\right)\] So, the charge flowing to conductor B is: \(Q_{flow} = Q\left(\frac{R}{R + r}\right)\) Thus, the correct answer is (A).