Chapter 17, Problem 76GP

To determine

The amount of charge lost.

Answer to Problem 76GP

  $4.4\times {10}^{-6}C$

Explanation of Solution

Given data:

Capacitance, $C=2.5\mu F=2.5\times {10}^{-6}F$

Potential, $V=6.0V$

Formula used:

The potential energy stored in a capacitor:

  $PE=\frac{1}{2}QV$ .

Since, $Q=CV$ , then

  $PE=\frac{1}{2}QV=\frac{1}{2}C{V}^2$

Where,

  $C$ is the capacitance.

  $\text{V}$ is the potential difference.

  $Q$ is the charge on each plate.

Calculation:

The energy stored in the capacitor: $PE=\frac{1}{2}QV=\frac{1}{2}C{V}^2$

The final energy is half the initial energy.

To calculate the amount of charge lost, first find the final charge and then subtract it form the initial charge:

  $\begin{array}{l}{E}_{final}+\frac{1}{2}{E}_{initial}\\ \Rightarrow \frac{1}{2}\frac{Q_{final}^2}{C}=\frac{1}{2}\frac{1}{2}\frac{Q_{initial}^2}{C}\\ \Rightarrow Q_{final}=\frac{1}{\sqrt{2}}{Q}_{initial}\end{array}$

  $Q_{lost}=Q_{initial}-Q_{final}$

  $=Q_{initial}\left(1-\frac{1}{\sqrt{2}}\right)$

  $=CV\left(1-\frac{1}{\sqrt{2}}\right)\mathrm{...}\left(1\right)$

Substitute all the values in equation $\left(1\right)$

  $\begin{array}{l}=\left(2.5\times {10}^{-6}F\right)\left(6.0V\right)\left(0.2929\right)\\ =4.4\times {10}^{-6}C\end{array}$

Conclusion:

The required value is $4.4\times {10}^{-6}C$ .