Question

In the following figure, the particles have charges q1 = q2 = 100 nC and q3 =

− q4 = 200 nC, distance a = 5 cm. Find the electrostatic force on charge 4 in unit-

vector notation and magnitude-angle notation.

Answer 1

Answer

X component of force can be written as

$F_x=\frac{1}{\sqrt{2}}K\frac{q_1q_4}{(5\sqrt{2}cm)^2}+K\frac{q_2q_4}{(5cm)^2}$

$F_x=\frac{1}{\sqrt{2}}(9\times10^9)\frac{(100nc)(200nc)}{(5\sqrt{2}cm)^2}+(9\times10^9)\frac{(100nc)(200nc)}{(5cm)^2}$

$F_x=0.05\hat i N$

Y component can be given as

$F_y=\frac{1}{\sqrt{2}}(9\times10^9)\frac{(100nc)(200nc)}{(5\sqrt{2}cm)^2}+(9\times10^9)\frac{(200nc)(200nc)}{(5cm)^2}$

$F_y=0.16\hat{j}N$

Angle with horizontal can be given as

$cos\theta=\frac{F_x}{F_R}=\frac{0.05}{0.167}\\\theta=72.6°$