Chapter 21.2, Problem 37PP

To determine

The capacitor which has a greater larger potential difference across it and the magnitude of that potential difference.

Answer to Problem 37PP

1^st capacitor will have larger potential difference across it with the magnitude of 106 V.

Explanation of Solution

Given:

The capacitance of the 1^st capacitor, $C_1=3.3\mu \text{F}$

The capacitance of the 2^nd capacitor, $C_2=6.8\mu \text{F}$

The charge across the capacitors, $q=3.5\times {10}^{-4}\text{C}$

Formula used:

The capacitance is given by,

  $C=\frac{q}{\Delta V}$

Where,

  $\Delta V$ is the potential difference and q is the charge stored in the capacitor.

Calculation:

The potential difference across 1^st capacitor will be,

  $\begin{array}{c}q=C_1\Delta V_1\\ 3.5\times {10}^{-4}=\left(3.3\times {10}^{-6}\right)\Delta V_1\\ \Delta V_1\approx 106\text{V}\end{array}$

And, the potential difference across 2^nd capacitor will be,

  $\begin{array}{c}q=C_2\Delta V_2\\ 3.5\times {10}^{-4}=\left(6.8\times {10}^{-6}\right)\Delta V_2\\ \Delta V_2\approx 51.47\text{V}\end{array}$

Conclusion:

Hence, the potential difference across 1^st capacitor with capacitance $3.3\mu \text{F}$ is greater than the charge on the 2^nd capacitor.