# What is the magnitude of the electric field at a point 0.0075 m from a 0.0035 C charge?

Nov 23, 2015

$5.6 \times {10}^{11} \text{N/C}$

#### Explanation:

The electric field strength is the force per unit charge.

It is given by:

$F = \frac{1}{4 \pi {\epsilon}_{0}} . \frac{Q}{r} ^ 2$

$\frac{1}{4 \pi {\epsilon}_{0}}$ is a constant and has the value $9 \times {10}^{9} \text{m/F}$

$\therefore F = 9 \times {10}^{9} \times \frac{0.0035}{{0.0075}^{2}}$

$\therefore F = 5.6 \times {10}^{11} \text{N/C}$