Question

Between the plates of a parallel plate capacitor of plate area A and capacity 0.025µF, a metal plate of area, A and thickness equal to $\frac{1}{3}$ of the separation between the plates of the capacitor is introduced. If the capacitor is charged to 100 V and battery is removed, then the amount of work done to remove the metal plate from the capacitor is

• A. 4.17 x 10⁻⁵ J

Answer 1

Answer: (A)

4.17 x 10⁻⁵ J

Answer 2

1. The original capacitor has capacitance

   $$C_0=0.025\mu F.$$

2. The capacitor is charged to $100\,V$ (battery removed later), so the charge is

   $$Q=C_0\,V=0.025\times10^{-6}\,F\times 100\,V=2.5\times10^{-6}\,C.$$

3. A metal plate of thickness $t=\frac{d}{3}$ is fully inserted between the plates (but does not contact them). This reduces the effective distance for the dielectric (air) to

   $$d_{\text{eff}}=d-t=d-\frac{d}{3}=\frac{2d}{3}.$$

   Thus, the new capacitance becomes

   $$C'=\frac{\epsilon_0\,A}{d_{\text{eff}}}=\frac{\epsilon_0\,A}{\frac{2d}{3}}=\frac{3}{2}\,\frac{\epsilon_0\,A}{d}=\frac{3}{2}\,C_0.$$

   Numerically,

   $$C'=\frac{3}{2}\times0.025\mu F=0.0375\mu F.$$

4. With the battery disconnected, the charge remains constant.

   • Energy when the metal plate is inserted:

     $$U_{\text{in}}=\frac{Q^2}{2\,C'}=\frac{Q^2}{2\times\frac{3}{2}\,C_0}=\frac{Q^2}{3\,C_0}.$$

   • Energy when the plate is removed (i.e. original capacitor):

     $$U_{\text{final}}=\frac{Q^2}{2\,C_0}.$$

5. The work done in removing the plate (against the attractive force) equals the increase in stored energy:

   $$W=\Delta U=U_{\text{final}}-U_{\text{in}}=\frac{Q^2}{2\,C_0}-\frac{Q^2}{3\,C_0}=\frac{Q^2}{6\,C_0}.$$

6. Substitute the known values:

   $$Q^2=(2.5\times10^{-6})^2=6.25\times10^{-12}\,C^2,$$ $$C_0=0.025\times10^{-6}\,F.$$

   Then,

   $$W=\frac{6.25\times10^{-12}}{6\times0.025\times10^{-6}}=\frac{6.25\times10^{-12}}{1.5\times10^{-7}}=4.17\times10^{-5}\,J.$$