Question:

A $ 400\,pF $ capacitor is charged by a $ 100\,V $ supply. How much electrostatic energy is lost in the process of disconnecting from the supply and connecting another uncharged $ 400\,pF $ capacitor?

Updated On: Jun 23, 2023
  • $ 10^{-5}\,\,J $
  • $ 10^{-6}\,\,J $
  • $ 10^{-7}\,\,J $
  • $ 10^{-4}\,\,J $
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The Correct Option is B

Solution and Explanation

$C=400 \times 10^{-12} F$
$V=100\, V$
Initially energy stored $=\frac{1}{2} C V^{2}$
$=\frac{1}{2} 400 \times 10^{-12} \times\left(10^{2}\right)^{2}$
$=\frac{1}{2} \times 4 \times 10^{-10} \times 10^{4}$
$=2 \times 10^{-6} J$
$\Rightarrow Q=C V=4 \times 10^{-10} \times 100$
$=4 \times 10^{-8} C$
$\Rightarrow$ By again connecting the charged capacitor with a uncharged ideal capacitor.
$\Rightarrow$ Now, the charge $Q$ will divide into two capacitors.
$\Rightarrow$ As $V$ and $C$ are same for both, $Q'$ will be same for both,
$\Rightarrow$ So $Q'=\frac{Q}{2}=2 \times 10^{-8} C$
$\Rightarrow$ So, energy stored in both the capacitors $=\left(\frac{\left(Q'\right)^{2}}{2 C}\right)=\frac{\left(Q'\right)^{2}}{2 C}$
$=\frac{\left(2 \times 10^{-8}\right)}{8 \times 10^{-10}}=10^{-6} J$
So energy lost $=2 \times 10^{-6}-0.25 \times 10^{-6}$
$=0.75 \times 10^{-6} J \approx 10^{-6} J$
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Concepts Used:

Electrostatic Potential and Capacitance

Electrostatic Potential

The potential of a point is defined as the work done per unit charge that results in bringing a charge from infinity to a certain point.

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In Series

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In Parallel

Both Capacitor C1 and C2 are connected in parallel. When the capacitors are connected parallelly then the total capacitance that is Ctotal is any one of the capacitor’s capacitance.