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*10^(-8))\vec{E} \\ \vec{E}=(0.060)/(-2*10^(-8))\hat{\imaginaryI} \\ \vec{E}=-3*10^6\hat{\imaginaryI} \end{gathered}`

Therefore, the magnitude of the electric field is

`3*10^6\text{ }(N)/(C)`

and it points to the left.