Chapter 17, Problem 50P

(a)

To determine

The factor by which the energy stored in the electric field changed.

Answer to Problem 50P

Twice the original energy stored.

Explanation of Solution

Given:

The parallel plate capacitor has fixed charges $\text{+Q}$ and $\text{-Q}$ with each plate area to be $\text{A}$ .

Formula used:

  $\text{C=}\frac{{\text{ε}}_{\text{0}}\text{A}}{\text{d}}$

Where,

  ${\text{ε}}_{\text{0}}$ is the vacuum permittivity.

  $\text{A}$ is the plate area.

  $\text{d}$ is the distance between the plates.

Moreover, the energy stored in the capacitor is

  $\text{U=}\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}$

Where,

  $\text{Q:}$ is the charge stored in the capacitor.

Calculation:

The distance between the capacitor plates is doubled.

  $\text{C=}\frac{{\text{ε}}_{\text{0}}\text{A}}{\text{d}}$

  ${\text{C}}^{\text{'}}\text{=}\frac{{\text{ε}}_{\text{0}}\text{A}}{\text{2d}}$

  ${\text{C}}^{\text{'}}\text{=}\frac{\text{C}}{\text{2}}$

The change in the energy stored is

  $\text{U=}\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}$

  ${\text{U}}^{\text{'}}\text{=}\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\frac{{\text{C}}^{\text{'}}}{\text{2}}}$

  ${\text{U}}^{\text{'}}\text{=2U}$

(b)

To determine

The work done in doubling the distance between the plates.

Answer to Problem 50P

  $\frac{\text{1}}{\text{2}}\left(\frac{{\text{dQ}}^2}{{\text{ε}}_{\text{0}}\text{A}}\right)$ .

Explanation of Solution

Given:

The parallel plate capacitor has fixed charges $\text{+Q}$ and $\text{-Q}$ with each plate area to be $\text{A}$ .

Formula used:

The capacitance for an air field capacitor is

  $\text{C=}\frac{{\text{ε}}_{\text{0}}\text{A}}{\text{d}}$

Where,

  ${\text{ε}}_{\text{0}}$ is the vacuum permittivity.

  $\text{A}$ is the plate area.

  $\text{d}$ is the distance between the plates.

Moreover, the energy stored in the capacitor is

  $\text{U=}\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}$

Where,

  $\text{Q}$ is the charge stored in the capacitor.

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

The work done to separate the plates is ${\text{U}}^{\text{'}}\text{-U}$ .

Calculation:

The change in enrgy is the subtraction of the final energy stored in the capacitor to the initial energy stored in the capacitor.

  $\Delta {\text{W=U}}^{\text{'}}\text{-U}$

  $\Delta \text{W=}\frac{\text{1}}{\text{2}}\frac{{\left({\text{Q}}^{\text{'}}\right)}^{\text{2}}}{{\text{C}}^{\text{'}}}\text{-}\frac{\text{1}}{\text{2}}\frac{{\text{Q}}^{\text{2}}}{\text{C}}$

  $\Delta \text{W=}\frac{\text{1}}{\text{2}}\left(\frac{{\text{dQ}}^2}{{\text{ε}}_{\text{0}}\text{A}}\right)$