Question

A negative charge of 2.0×10−8 C experiences a force of 0.060 N to the right in an electric field. What is the magnitude and direction of the field?

Answer 1

The force exerted on a charge q by an electric field is given by:

$\vec{F}=q\vec{E}$

We know that the force is exerted to the right which means the force is positive and pointing in the direction of the unit vector i, then we have:

$\begin{gathered} 0.060\hat{\imaginaryI}=(-2\times10^{-8})\vec{E} \\ \vec{E}=\frac{0.060}{-2\times10^{-8}}\hat{\imaginaryI} \\ \vec{E}=-3\times10^6\hat{\imaginaryI} \end{gathered}$

Therefore, the magnitude of the electric field is

$3\times10^6\text{ }\frac{N}{C}$

and it points to the left.