## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 16 - Problems - Page 615: 48

#### Answer

A charge of $~-0.432~q~$ must be placed at the point $(4.0, 0.0)$.

#### Work Step by Step

First, we can find the magnitude of the electric field at point $(4.0, 3.0)$ due to the two charges $q$. By symmetry, the horizontal component of the electric field due to the two charges $q$ is zero. Let $\theta$ be the angle above the horizontal of the line from each point charge $q$ to the point $(4.0,3.0)$. The vertical component of the electric field at the point $(4.0,3.0)$ is the sum of the vertical components of the electric field due to each point charge $q$: $E_y = 2\times \frac{k~q}{r^2}~(sin~\theta)$ $E_y = 2\times \frac{k~q}{(5.0~m)^2}~(\frac{3.0}{5.0})$ $E_y = 0.048~k~q$ The third charge at the point $(4.0, 0.0)$ must be negative and the electric field due to this point charge must be equal in magnitude to $E_y$. We can find $Q$, the magnitude of the charge at the point $(4.0, 0.0)$: $\frac{k~Q}{d^2} = E_y$ $\frac{k~Q}{d^2} = 0.048~k~q$ $Q = 0.048~q~d^2$ $Q = 0.048~q~(3.0~m)^2$ $Q = 0.432~q$ A charge of $~-0.432~q~$ must be placed at the point $(4.0, 0.0)$.

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