91Ó°ÊÓ

$Q_1$= 80.0 nC is at (0, 3.00 mm)

$Q_2$is at (0, -3.00 mm)

$Q_3$= 18.0 nC is at (4.00 mm, 0)

Use Coulomb’s law to solve the problem.

Charge$Q_1=80.0×{10}^{−9}C$is on the y axis at y = 0.003 m, and charge

$Q_2=80.0×{10}^{−9}C$is on the y axis at y = –0.003 m.

The force on particle 3 (which has a charge of$q=18.0×{10}^{−9}C$is due to the vector sum of the repulsive forces from Q1 and Q2.

In symbols,$F_{31}+F_{32}=F_3$, where

$F_{31}=k\frac{q_3\left|q_1\right|}{r_{31}^2},F_{32}=k\frac{q_3\left|q_2\right|}{r_{32}^2}$

Using the Pythagorean theorem, we have$r_{31}$= $r_{32}$= 0.005 m.

In magnitude-angle notation (particularly convenient if one uses a vector-capable calculator in polar mode), the indicated vector addition becomes

$F_3=(0.518∠−37°)+(0.518∠37°)=(0.829∠0°)$

$⇒F_3=(0.829N)\widehat{i}$

Switching the sign of Q2 amounts to reversing the direction of its force on q.

Consequently, we have

$F_3=(0.518∠−37°)+(0.518∠−143°)=(0.621∠−90°)$

$F_3=−(0.621N)\widehat{j}$