Chapter 16, Problem 25P

To determine

To find: the magnitude and direction of the electric field

Answer to Problem 25P

The direction of the electric field is in upward direction since the charge is negative.

The magnitude of the electric field $=-9.54\times {10}^5\text{N}/\text{C}$

Explanation of Solution

Given:

A downward force of $8.4\text{N}$ is exerted on a $-8.8\mathrm{\mu }\text{C}$

Formula used:

The electric field at the point is expressed as follows:

  $\mathrm{E}=\frac{{\mathrm{F}}_{\mathrm{e}}}{\mathrm{q}}$

Here, ${\mathrm{F}}_{\mathrm{e}}$ is the force of the charge and $\mathrm{q}$ is the charge.

Calculation:

Substitute $8.4\text{N}$ for ${\mathrm{F}}_{\mathrm{e}}$ and $-8.8\mathrm{\mu }\text{C}$ for $\mathrm{q}$

  $\begin{array}{l}\mathrm{E}=\frac{8.4\text{N}}{(-8.8\mathrm{\mu }\text{C})\left(\frac{{10}^{-6}\text{C}}{1\mathrm{\mu }\text{C}}\right)}\\ =-9.54\times {10}^5\text{N}/\text{C}\end{array}$

Therefore, the magnitude of the electric field is $9.54\times {10}^5\text{N}/\text{C}$

Hence, the direction of the electric field is in upward direction since the charge is negative.