Chapter 22, Problem 1P

To determine

The rate at which the electric field is changing between the plates if the square plates are $1.60cm$ on a side.

Answer to Problem 1P

The answer is $7.94\times {10}^{14}V/ms$

Explanation of Solution

Given info:

Current, $I=1.8A$

Permittivity of free space, ${\epsilon }_0=8.85\times {10}^{-12}F/M$

Formula used:

It is known that:

  $I_D={\epsilon }_{0}A\left(\frac{\Delta E}{\Delta t}\right)$

Where,

  $I_D$ is drawn current.

  $A$ is area,

  $\Delta t$ is time change

  $\Delta E$ is change in magnitude of electric field

  ${\epsilon }_0$ is permittivity of free space

Calculation:

Substitute given and known values in formula, $I_D={\epsilon }_{0}A\left(\frac{\Delta E}{\Delta t}\right)$

  $1.8A=\left(8.85\times {10}^{-12}F/M\right){\left(0.0190m\right)}^2\left(\frac{\Delta E}{\Delta t}\right)=7.94\times {10}^{14}V/ms$

Conclusion: The required rate is $7.94\times {10}^{14}V/ms$ .