## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 16 - Problems - Page 615: 36

#### Answer

The magnitude of the electric field at point $S$ is $~\frac{7~k~q}{4~d^2}~$ and the electric field is directed in the negative x-direction.

#### 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 $q$: $E_1 = \frac{k~q}{(2d)^2} = \frac{k~q}{4~d^2}$ 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 $2q$: $E_2 = \frac{k~(2q)}{d^2} = \frac{2~k~q}{d^2}$ We can find the net electric field due to both point charges: $E = E_1-E_2$ $E = \frac{k~q}{4~d^2}-\frac{2~k~q}{d^2}$ $E = -\frac{7~k~q}{4~d^2}$ The magnitude of the electric field at point $S$ is $~\frac{7~k~q}{4~d^2}~$ and the electric field is directed in the negative x-direction.

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