Question

(a) a 900pF capacitor is charged by 100V battery. how much electrostatic energy is stored by the capacitor? (b) the capacitor is disconnected from the battery and connected to another 900pF capacitor. what us the electrostatic energy stored by the system?

Answer 1

a)
Given
C = 900pF, V = 100 V
E= 1/2 CV² = 4.5 * 10^-6 J
b)
Let the common potential be V
Using law of conservation of charge,
CV = (C+C)V'
V'= 50 V
Therefore,
E'= 1/2(C+C)V'²
= 2.25 * 10^-6 J

Answer 2

Given:

Charged voltage = 100 V battery in 900 pF capacitor  

Solution:

a. Electrostatic energy stored in capacitor.

Stored electrostatic energy by the capacitor is given by the formula:

$E=\frac{1}{2} c v^{2}$

$=\frac{1}{2} \times 900 \times 10^{-12} \times 100^{2}$

$E=4.5 \times 10^{-6} \ J$

Where,

E – Electrostatic energy

c – Charge = $900 \times 10^{-12} \ \mathrm{C}$

v - Potential = 100 V

b. System’s electrostatic energy that is stored when the capacitor is connected to another capacitor whose value is 900pF and disconnected from battery.

$\frac{1}{c^{\prime}}=\frac{1}{c}+\frac{1}{c}$

Here, c’ represents total charge in the capacitor.

$\Rightarrow \frac{1}{c^{\prime}}=\frac{1}{900 \times 10^{-12}}+\frac{1}{900 \times 10^{-12}}=\frac{2}{900 \times 10^{-12}}$

$\Rightarrow c^{\prime}=\frac{900 \times 10^{-12}}{2}$

$\therefore c^{\prime}=450 \times 10^{-12} \ F$

Thus, the stored energy in the system is given as:

$E^{\prime}=\frac{c^{\prime}}{2} \times v^{2}$

$E^{\prime}=\frac{450 \times 10^{-12}}{2} \times 100^{2}$

$\Rightarrow E^{\prime}=2.25 \times 10^{-6} \ J$