Chapter 16, Problem 24P

To determine

To find: the magnitude and direction of the electric field

Answer to Problem 24P

The direction of the electric field is towards south.

The magnitude of the electric field $=2.34\times {10}^5\text{N}/\text{C}$

Explanation of Solution

Given:

Force exerted on the proton towards south, $\mathrm{F}=3.75\times {10}^{-14}\text{N}$

Formula used:

The magnitude of the electric field at any point is defined as the ratio of the force exerted on a tiny positive charge to the magnitude of the test charge

  $\mathrm{E}=\frac{\mathrm{F}}{\mathrm{Q}}\dots \dots (1)$

Here,

F is force exerted on the particle

Q is charge of the particle

Calculation:

Charge of proton, $\mathrm{Q}=1.6\times {10}^{-19}\text{C}$

The magnitude of the electric field is,

  $\begin{array}{l}\mathrm{E}=\frac{\mathrm{F}}{\mathrm{Q}}\\ =\frac{3.75\times {10}^{-14}\text{N}}{1.6\times {10}^{-19}\text{C}}\\ =2.34\times {10}^5\text{N}/\text{C}\end{array}$

The direction of the electric field is always from positive to negative, so in this case, the direction of the electric field is towards south.