Chapter 21, Problem 95A

To determine

Work done to move the charge.

Answer to Problem 95A

  $W=6.4\times {10}^{-6}\text{J}$

Explanation of Solution

Given:

Charge, $q=0.25\mu \text{C}$

Distance between the parallel plates, $d=0.4\text{cm}=0.004\text{m}$

Electric field between the plates, $E=6400\text{N/C}$

Formula Used:

The work done in moving the charge is,

  $W=q\Delta V=qEd\left(\text{Since, }\Delta V=Ed\right)$

Where, $q$ is the charge on the capacitor, $\Delta V$ electric potential difference, $E$ is Electric field intensity and $d$ is distance between the parallel plates.

Calculation:

On putting the values in above relation,

  $\begin{array}{c}W=qEd\\ =0.25\times {10}^{-6}\times 6400\times 0.004\\ =6.4\times {10}^{-6}\text{J}\end{array}$

Conclusion:

Therefore, the work done in moving the charge is $W=6.4\times {10}^{-6}\text{J}$