Chapter 17, Problem 36P

To determine

The capacitance of the capacitor.

Answer to Problem 36P

The answer is $7.5\times {10}^{-7}F$ .

Explanation of Solution

Given:

Charge on a capacitor increases by $18\mu C=18\times {10}^{-6}C$

Voltage across charge increases from $97V$ to $121V$

Formula used:

It is known that $Q=CV$ , here the amount of charge $Q$ acquired by each plate is proportional to the magnitudes of potential difference $V$ b/w them.

Calculation:

Suppose $Q_1$ and $V_1$ be the initial charge and voltage on the capacitor.

Suppose $Q_2$ and $V_2$ be the final charge and voltage on the capacitor.

To relate the charges and voltages to the capacitance use expression $Q=CV$ ;

  $Q_1=C{V}_1$ and $Q_2=C{V}_2$

  $Q_2-Q_1=C{V}_1-C{V}_2=C\left(V_2-V_1\right)\to C=\frac{Q_2-Q_1}{V_2-V_1}=\frac{18\times {10}^{-6}C}{24V}=7.5\times {10}^{-7}F$

Conclusion:

The required capacitance is $7.5\times {10}^{-7}F$ .