Question

A hollow spherical conductor of radius R has a charge Q on it. A small dent on the surface decreases the volume of the spherical conductor by 2%. Assume that the charge density on the surface does not change due to the dent and the electric field in the dent region remains same as other points on the surface. (a) AE is the electrostatic energy stored in the electric field in the shallow dent region and E is the total electrostatic energy of the spherical shell. Find the ΔΕ ratio E (b) Using the ratio obtained in part (a) calculate the percentage change in capacitance of the sphere due to the dent.

Answer 1

0.02 and -0.67%

Answer 2

Explanation of the solution:

(a) Find the ratio $\frac{\Delta E}{E}$: The original volume of the spherical conductor is $V_0 = \frac{4}{3}\pi R^3$. The dent decreases the volume by 2%, which means the change in volume is $\Delta V = 0.02 V_0$. The problem defines $\Delta E$ as the electrostatic energy stored in the electric field in the shallow dent region and $E$ as the total electrostatic energy of the spherical shell. Assuming that the energy associated with the dent region is proportional to the volume change it causes, relative to the total energy being proportional to the original volume, we can write the ratio as: $\frac{\Delta E}{E} = \frac{\Delta V}{V_0}$ Given that the volume decreases by 2%, $\frac{\Delta V}{V_0} = 0.02$. Therefore, $\frac{\Delta E}{E} = 0.02$.

(b) Calculate the percentage change in capacitance: The total electrostatic energy of a conductor with charge Q and capacitance C is given by $E = \frac{Q^2}{2C}$. Let the original capacitance be $C_0$ and the original energy be $E_0$. After the dent is made, the capacitance becomes $C'$ and the energy becomes $E'$. Assuming the charge Q on the conductor remains constant, we have $E_0 = \frac{Q^2}{2C_0}$ and $E' = \frac{Q^2}{2C'}$. The change in energy is $E' - E_0 = \frac{Q^2}{2C'} - \frac{Q^2}{2C_0} = \frac{Q^2}{2} \left(\frac{1}{C'} - \frac{1}{C_0}\right)$. Let the change in capacitance be $\Delta C = C' - C_0$. For a small dent, $\Delta C$ is small compared to $C_0$. $E' - E_0 = \frac{Q^2}{2} \left(\frac{C_0 - C'}{C_0 C'}\right) = \frac{Q^2}{2} \frac{-\Delta C}{C_0 (C_0 + \Delta C)}$. For small $\Delta C$, $C' \approx C_0$, so $E' - E_0 \approx \frac{Q^2}{2C_0^2} (-\Delta C) = - \left(\frac{Q^2}{2C_0}\right) \frac{\Delta C}{C_0} = -E_0 \frac{\Delta C}{C_0}$. So, $\frac{E' - E_0}{E_0} = -\frac{\Delta C}{C_0}$.

The change in total energy is also related to the work done to deform the conductor against the electrostatic pressure. The electrostatic pressure just outside the surface of a conductor is $P = \frac{\sigma^2}{2\epsilon_0}$. The work done on the conductor to reduce its volume by $\Delta V$ is $W = P \Delta V$. This work increases the electrostatic energy of the conductor, so $E' - E_0 = P \Delta V$. For a spherical conductor, the total energy can be related to the pressure and volume: $E_0 = \frac{Q^2}{8\pi\epsilon_0 R} = \frac{(\sigma 4\pi R^2)^2}{8\pi\epsilon_0 R} = \frac{16\pi^2 \sigma^2 R^4}{8\pi\epsilon_0 R} = \frac{2\pi \sigma^2 R^3}{\epsilon_0}$. Since $V_0 = \frac{4}{3}\pi R^3$, we have $R^3 = \frac{3V_0}{4\pi}$. $E_0 = \frac{2\pi \sigma^2}{\epsilon_0} \left(\frac{3V_0}{4\pi}\right) = \frac{3 \sigma^2 V_0}{2\epsilon_0}$. The pressure is $P = \frac{\sigma^2}{2\epsilon_0}$. So, $E_0 = 3 P V_0$, or $P = \frac{E_0}{3V_0}$. The energy change is $E' - E_0 = P \Delta V = \frac{E_0}{3V_0} \Delta V = \frac{1}{3} E_0 \frac{\Delta V}{V_0}$. So, $\frac{E' - E_0}{E_0} = \frac{1}{3} \frac{\Delta V}{V_0}$.

Equating the two expressions for $\frac{E' - E_0}{E_0}$: $-\frac{\Delta C}{C_0} = \frac{1}{3} \frac{\Delta V}{V_0}$. We are given $\frac{\Delta V}{V_0} = 0.02$. $\frac{\Delta C}{C_0} = -\frac{1}{3} \times 0.02 = -\frac{0.02}{3}$. The percentage change in capacitance is $\frac{\Delta C}{C_0} \times 100\% = -\frac{0.02}{3} \times 100\% = -\frac{2}{3}\%$. As a decimal, $-\frac{2}{3}\% \approx -0.666...\%$. Rounded to two decimal places, this is $-0.67\%$. The capacitance decreases due to the dent.

Answer:

(a) The ratio $\frac{\Delta E}{E}$ is $0.02$. (b) The percentage change in capacitance is approximately $-0.67\%$.

Subject: Physics Chapter: Electrostatic Potential and Capacitance Topic: Electrostatic energy of a conductor, Capacitance of a spherical conductor, Effect of shape on capacitance

Difficulty: Medium Question Type: Descriptive