Three charged particles form a triangle: particle 1 with charge$Q1=80.0nC$is at x ycoordinates ($0,3.00mm$), particle 2 with charge$Q2$is at ($0,-3.00mm$), and particle 3 with charge$q=18.0nc$is at ($4.00mm,0$). In unit-vector notation, what is the electrostatic force on particle 3 due to the other two particles if$Q_2$is equal to (a)$80.0nC$and (b)$-80.0nC$?

$Q_1$= 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$Q_1=80.0\times {10}^{-9}C$is on the y axis at y = 0.003 m, and charge

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

The force on particle 3 (which has a charge of$q=18.0\times {10}^{-9}C$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

$F_3=(0.518\angle -37°)+(0.518\angle 37°)=(0.829\angle 0°)$

$\Rightarrow F_3=(0.829N)\widehat{i}$

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

Consequently, we have

$F_3=(0.518\angle -37°)+(0.518\angle -143°)=(0.621\angle -90°)$

$F_3=-(0.621N)\widehat{j}$