Question

Derive an expression for energy stored in a charged capacitor. A spherical metal ball of radius 15 cm carries a charge of 2μC. Calculate the electric field at a distance of 20 cm from the center of the sphere.

Hint:

The energy stored in a capacitor is directly proportional to the capacitance and the square of the voltage. The electric field outside a spherical charge distribution behaves as if all the charge were concentrated at the center of the sphere.

Answer 1

Energy Stored in a Charged Capacitor
The energy \( U \) stored in a charged capacitor is given by the formula:
\[ U = \frac{1}{2} C V^2 \] where: - \( C \) is the capacitance of the capacitor,
- \( V \) is the voltage across the capacitor.
Capacitance of a Spherical Capacitor:
For a spherical capacitor (or a sphere with charge), the capacitance \( C \) is given by:
\[ C = 4 \pi \epsilon_0 R \] where: - \( \epsilon_0 = 8.85 \times 10^{-12} \, {F/m} \) is the permittivity of free space,
- \( R \) is the radius of the sphere.
Given \( R = 0.15 \, {m} \), we can calculate the capacitance:
\[ C = 4 \pi \times 8.85 \times 10^{-12} \times 0.15 = 1.67 \times 10^{-12} \, {F}. \]
Voltage across the Spherical Capacitor:
The voltage \( V \) across the capacitor is related to the charge \( Q \) and capacitance \( C \) by the formula:
\[ V = \frac{Q}{C} \] Given the charge \( Q = 2 \, \mu C = 2 \times 10^{-6} \, {C} \), we can calculate the voltage:
\[ V = \frac{2 \times 10^{-6}}{1.67 \times 10^{-12}} = 1.20 \times 10^6 \, {V}. \] 
Energy Stored in the Capacitor:
Now, we can calculate the energy stored in the capacitor using the formula \( U = \frac{1}{2} C V^2 \):
\[ U = \frac{1}{2} \times (1.67 \times 10^{-12}) \times (1.20 \times 10^6)^2 \] \[ U = \frac{1}{2} \times 1.67 \times 10^{-12} \times 1.44 \times 10^{12} \] \[ U = 1.20 \, {J}. \] Thus, the energy stored in the charged capacitor is 1.20 joules.
Electric Field at a Distance of 20 cm from the Center of the Sphere:
The electric field \( E \) at a distance \( r \) from the center of a spherical charge distribution is given by:
\[ E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2} \] where:
- \( Q = 2 \times 10^{-6} \, {C} \) is the charge on the sphere,
- \( r = 0.20 \, {m} \) is the distance from the center of the sphere (20 cm).
Substituting the values:
\[ E = \frac{9 \times 10^9 \times 2 \times 10^{-6}}{(0.2)^2} = \frac{18 \times 10^3}{0.04} \] \[ E = 4.5 \times 10^5 \, {N/C}. \] Thus, the electric field at a distance of 20 cm from the center of the sphere is \( 4.5 \times 10^5 \, {N/C} \).