Question

A $6 μF$ capacitor is charged by a $300 V$ supply. It is then disconnected from the supply and is connected to another uncharged $3 μF$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ?

A $4 μF$ capacitor is charged by a $200 V$ supply. It is then disconnected from the supply and is connected to another uncharged $2 μF$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

Accepted Answer

Given,
The capacitance of the charged capacitor is, $C_{1} = 6 μF = 6 {\times10}^{- 6} F$
The supply voltage is, $200 V$
The electrostatic energy stored in the charged capacitor is,
$E_{1} = \frac{1}{2} C_{1} V_{1}^{2} = \frac{1}{2} (6 \times 10^{- 6}) {(300)}^{2} = 27 \times 10^{- 2} J$
The capacitance of the uncharged capacitor is,
$C_{2} = 3 μF = 3 {\times10}^{- 6} F$
When the charged capacitor is connected to an uncharged one, say the voltage across both the capacitors become $V_{2}$ From the conservation of charge, the initial charge in the capacitor is equal to the sum of the charges across both the capacitors. Hence,
$C_{1} V_{1} = (C_{1} + C_{2}) V_{2}$
$\begin{matrix} V_{2} = \frac{C_{1}}{(C_{1} + C_{2})} V_{1} = \frac{6 \times 10^{- 6}}{(6 \times 10^{- 6} + 3 \times 10^{- 6})} \times 300 \end{matrix} = 2 \times 10^{2} V$
Therefore, the electrostatic potential energy stored in the combination of capacitors is,
$E_{2} = \frac{1}{2} (C_{1} + C_{2}) V_{2}^{2} = \frac{1}{2} (6 \times 10^{- 6} + 3 \times 10^{- 6}) {(2 \times 10^{2})}^{2} = 18 \times 10^{- 2} J$
Therefore, the energy lost by charged capacitor is,
$Δ E = E_{1} - E_{2} = 27 \times 10^{- 2} J - 18 \times 10^{- 2} J = 9 \times 10^{- 2} J$
Thus, the energy lost as heat and electromagnetic radiation is $9 {\times10}^{- 2} J$ .

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