Chapter 17, Problem 61GP

To determine

The amount of charge that can be placed on a parallel-plate capacitor

Explanation of Solution

Given data:

Area, $A=56c{m}^2=56\times {10}^{-4}{m}^2$

Electric field, $E=3.0\times {10}^{6}V/m$

Formula used:

  $Q=CV$

Where,

  $C$ is the capacitance of a capacitor.

  $\text{V}$ is the magnitude of the change in potential across the capacitor.

  $Q$ is magnitude of charge on each plate of the capacitor.

Calculation:

Use the formula $Q=CV$ to find the amount of charge that can be placed on a parallel-plate capacitor.

  $\begin{array}{c}Q=CV\\ =\frac{{\text{ε}}_{0}A}{d}V\\ ={\text{ε}}_{0}A\frac{V}{d}\\ ={\text{ε}}_{0}AE\mathrm{...}\left(1\right)\end{array}$

Where,

  $A$ is the area.

  $E$ is the electric field.

  ${\text{ε}}_{\text{0}}$ is the permittivity of free space.

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

  $\begin{array}{l}=\left(8.85\times {10}^{-12}{C}^2/N\cdot m^2\right)\left(56\times {10}^{-4}{m}^2\right)\left(3.0\times {10}^{6}V/m\right)\\ =1.5\times {10}^{-7}C\end{array}$

Conclusion:

The required value is $1.5\times {10}^{-7}C$ .