Problem 149 - A Complete Resource Book in Physics for JEE Main

Question

A spherical shell of radius \(R_{1}=10 \mathrm{~cm}\) with a uniform charge \(q=6 \mu \mathrm{C}\) has a point charge \(q_{0}=3 \mu \mathrm{C}\) at its centre. Find the work performed by electric forces in milli joules during the shell expansion from radius \(R_{1}\) to radius \(R_{2}=20 \mathrm{~cm}\). Take \(\frac{1}{4 \pi \varepsilon_{0}}=9 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\).

Step-by-Step Solution

Answer

The work performed by electric forces during the shell expansion from radius \(R_1\) to radius \(R_2\) is 540 millijoules.

Step 1: Find the electric field at the surface of the shell due to the point charge at its center.

To find the electric field at the surface of the shell due to the point charge at its center, we use the formula for the electric field of a point charge: \(E = \frac{1}{4 \pi \varepsilon_0} \frac{q_0}{R_1^2}\) We are given the values for \(q_0\), \(R_1\) and \(\frac{1}{4 \pi \varepsilon_{0}}\), so we can plug these in: \(E = \frac{9\times10^9 Nm^2/C^2 \times 3\times10^{-6}C}{(0.1m)^2}\) \(E = 2.7\times10^6 N/C\)

Step 2: Find the potential difference between the initial and final radii.

Next, we need to calculate the potential difference between the initial and final radii due to the electric field. We can use the following formula for the potential difference between two points in the electric field of a point charge: \(\Delta V = \frac{1}{4 \pi \varepsilon_0}\frac{q_0}{R_2} - \frac{1}{4 \pi \varepsilon_0}\frac{q_0}{R_1}\) Plugging in the given values for \(q_0\), \(R_1\), \(R_2\), and \(\frac{1}{4 \pi \varepsilon_{0}}\): \(\Delta V = \frac{9\times10^9 Nm^2/C^2\times 3\times10^{-6}C}{0.2m} - \frac{9\times10^9 Nm^2/C^2\times 3\times10^{-6}C}{0.1m}\) \(\Delta V = 9\times10^4 V\)

Step 3: Calculate the work done by the electric forces.

Finally, we can use the formula for the work done by the electric forces during the expansion of the shell: \(W = q\Delta V\) Plugging in the given values for \(q\) and the calculated value for \(\Delta V\): \(W = 6\times10^{-6}C\times 9\times10^4 V\) \(W = 5.4\times10^{-1} J\) Since the problem asks for the work done in millijoules (mJ), we can convert the work done to that unit: \(W = 5.4\times10^{-1} J\times \frac{10^3 mJ}{1 J}\) \(W = 540 mJ\) Hence, the work performed by electric forces during the shell expansion from radius \(R_1\) to radius \(R_2\) is 540 millijoules.