Chapter 16, Problem 47P

To determine

The charge on each plate.

Answer to Problem 47P

The charge on each plate is $1.15\times {10}^{-9}\text{C}$ .

Explanation of Solution

Given info:

The electric field between two square metal plates is $E=130\text{N/C}$ .

The side of the plate is $1.0\text{m}$ .

The plate is separated by $3.0\text{cm}$ .

Formula used:

From the Gauss law

  $\begin{array}{c}E=\frac{\sigma }{{\epsilon }_0}\\ =\frac{Q/A}{{\epsilon }_0}\\ Q={\epsilon }_{0}EA\end{array}$

Here, ${\epsilon }_0$ is the absolute permittivity, E is the electric field and A is area.

Calculation:

Substituting the given values

  $\begin{array}{c}Q=\left(8.85\times {10}^{-12}{\text{C}}^{\text{2}}{\text{/Nm}}^{\text{2}}\right)\left(130\text{N/C}\right)\left(1{\text{m}}^{\text{2}}\right)\\ =1.15\times {10}^{-9}\text{C}\end{array}$

Conclusion:

Thus, the charge on each plate is $1.15\times {10}^{-9}\text{C}$ .