Chapter 21, Problem 4STP

To determine

To Choose: The correct option.

Answer to Problem 4STP

B. $8.60\times {10}^{-19}\text{J}$

Explanation of Solution

Given:

The electric field strength between the plates, $E=125\text{N/C}$

The distance between the plates, $d=4.3\text{cm}$

Formula used:

In terms of potential difference $\left(\Delta V\right)$ , Electric field strength $\left(E\right)$ can be given as,

  $E=\frac{\Delta V}{d}$

Here, d is the distance moved by the charge.

And, the electric potential difference can be written as,

  $\Delta V=\frac{W_{\text{on }q\text{'}}}{q\text{'}}$

Where,

  $W_{\text{on }q\text{'}}\text{'}$ is the work done to move the charge.

Calculation:

Combining above two equations, it can be written as,

  $E=\frac{W_{\text{on }q\text{'}}}{dq\text{'}}$

It is known that the charge on a proton is, $q\text{'}=1.6\times {10}^{-19}\text{C}$

Then, using the above formula, the work done in moving a proton will be,

  $\begin{array}{c}125=\frac{W_{\text{on }q\text{'}}}{\left(4.3\times {10}^{-2}\right)\left(1.6\times {10}^{-19}\right)}\\ W_{\text{on }q\text{'}}=8.60\times {10}^{-19}\text{J}\end{array}$

Conclusion:

Hence, the work done in moving a proton is $8.60\times {10}^{-19}\text{J}$ .