Chapter 17, Problem 66GP

To determine

The potential difference and the charge across each capacitor.

Answer to Problem 66GP

The charge of each capacitor is $\frac{{\text{Q}}_{\text{0}}{\text{C}}_{\text{1}}}{{\text{C}}_{\text{1}}{\text{+C}}_{\text{2}}}$ and $\frac{{\text{Q}}_{\text{0}}{\text{C}}_{\text{2}}}{{\text{C}}_{\text{1}}{\text{+C}}_{\text{2}}}$ , and the potential difference is $\frac{{\text{Q}}_{\text{0}}}{{\text{C}}_{\text{1}}{\text{+C}}_{\text{2}}}$ .

Explanation of Solution

Given:

A capacitor with capacitance ${\text{C}}_{\text{1}}$ and charge ${\text{Q}}_{\text{0}}$ is connected in series with an uncharged capacitor ${\text{C}}_{\text{2}}$ .

Formula used:

The charge of a capacitor is given by

  $\text{Q=CV}$

Where,

  $\text{C}$ is the capacitance of the capacitor.

  $\text{V}$ is the voltage across the capacitor.

Calculation:

Let the potential difference be $\text{V}$ .

Here,

  ${\text{Q}}_{\text{0}}={\text{C}}_{\text{1}}\text{V}$

  ${\text{Q}}_{\text{2}}={\text{C}}_{\text{2}}\text{V}$

Therefore,

  ${\text{Q}}_{\text{0}}={\text{Q}}_{\text{1}}+{\text{Q}}_{\text{2}}$

  ${\text{Q}}_{\text{0}}\text{=}\left({\text{C}}_{\text{1}}{\text{+C}}_{\text{2}}\right)\text{V}$

  $\text{V=}\frac{{\text{Q}}_{\text{0}}}{\left({\text{C}}_{\text{1}}{\text{+C}}_{\text{2}}\right)}$

The charge on capacitor ${\text{C}}_{\text{1}}$ is

  ${\text{Q}}_{\text{1}}{\text{=C}}_{\text{1}}\text{V}$

  ${\text{Q}}_{\text{1}}{\text{=Q}}_{\text{0}}\frac{{\text{C}}_{\text{1}}}{{\text{C}}_{\text{1}}{\text{+C}}_{\text{2}}}$

The charge on the capacitor ${\text{C}}_{\text{2}}$ is

  ${\text{Q}}_{\text{2}}{\text{=C}}_{\text{2}}\text{V}$

  ${\text{Q}}_{\text{2}}{\text{=Q}}_{\text{0}}\frac{{\text{C}}_{\text{2}}}{{\text{C}}_{\text{1}}{\text{+C}}_{\text{2}}}$

Conclusion:

The charge of a capacitor is the multiplication of the capacitance and the potential difference.