Chapter 17, Problem 4P

To determine

The potential difference between the plates, and which plate is at the higher potential.

Answer to Problem 4P

The potential difference is $4656\text{V}$ , and plate B is at higher potential.

Explanation of Solution

Given info:

The charge of electron is $q=-1.6\times {10}^{-19}\text{C}$ .

Kinetic energy is $KE=7.45\times {10}^{-16}\text{J}$ .

Formula used:

Kinetic energy in accelerating the electron through the potential difference is,

  $KE=q\Delta V$

Calculation:

Substituting the given values, we get

  $\begin{array}{c}7.45\times {10}^{-16}\text{J}=\left(1.6\times {10}^{-19}\text{C}\right)\left(\Delta V\right)\text{J}\\ \Delta V=\frac{7.45\times {10}^{-16}\text{J}}{1.6\times {10}^{-19}\text{C}}\\ =4656\text{V}\end{array}$

The electron moves from low potential to high potential, so plate B is at higher potential.

Conclusion:

Thus, the potential difference is $4656\text{V}$ , and plate B is at higher potential.