# Chapter 16 - Problems - Page 615: 35

$E = \frac{k~q}{2~d^2}$

#### Work Step by Step

The electric field due to the point charge $q$ is directed in the positive x-direction. We can find the magnitude of the electric field due to the point charge $q$: $E_1 = \frac{k~q}{d^2}$ The electric field due to the point charge $2q$ is directed in the negative x-direction. We can find the magnitude of the electric field due to the point charge $2q$: $E_2 = \frac{k~(2q)}{(2d)^2} = \frac{k~q}{2~d^2}$ We can find the net electric field due to both point charges: $E = E_1-E_2$ $E = \frac{k~q}{~d^2}-\frac{k~q}{2~d^2}$ $E = \frac{k~q}{2~d^2}$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.