A $4 \mu \mathrm{F}$ capacitor is charged by a $\mathbf{2 0 0} \mathbf{~ V}$ supply. It is then disconnected from the supply and is connected to another uncharged $2 \mu \mathrm{F}$ capacitors. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Home » A capacitor is charged by a supply. It is then disconnected from the supply and is connected to another uncharged capacitors. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

CBSE | Class 12 | Electrostatic Potential and Capacitance | NCERT | Physics

A $4 \mu \mathrm{F}$ capacitor is charged by a $\mathbf{2 0 0} \mathbf{~ V}$ supply. It is then disconnected from the supply and is connected to another uncharged $2 \mu \mathrm{F}$ capacitors. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Solution:
Given:
Capacitance of the capacitor, $\mathrm{C}_{1}=4 \mu \mathrm{F}$

Voltage, $V_{1}=200 \mathrm{~V}$

Capacitance of the uncharged capacitor, $C_{2}=2 \mu \mathrm{F}$

As we know, Electrostatic energy stored in $\mathrm{C}_{1}$ is given as

$\mathrm{E}{1}=(1 / 2) \mathrm{C}{1} \mathrm{~V}_{1}^{2}$
Substituting the value of know parameters we have,

$=(1 / 2) \times 4 \times 10^{-6} \times(200)^{2}$

$=8 \times 10^{-2} \mathrm{~J}$

When $C_{1}$ is disconnected from the power supply and connected to $C_{2}$ , the voltage acquired by it is $V_{2}$ .

Following the conservation of energy principle, the initial charge on capacitor $C_1$ is equal to the final charge on capacitors $C_1$ and $C_2$ , and vice versa.