Question

Consider three point charges q1= 5.0 μC , q2 = -2.0 μC and q3 = 5.0 μC, located at the corners of a triangle which has a coordinate of (0, 0) , (0, a) and (a, a) respectively. If a = 0.10 m, find

 a) the resultant force on q3.

 b) the magnitude and the direction of the resultant force on q3 

Answer 1

We know that

$F_1=\frac{kq_1q_2}{r^2}$

a=0.10

$F_1=\frac{9\times10^9\times5\times10^{-6}\times5\times10^{-6}}{(a\sqrt{2})^2}$

$F_1=\frac{0.1125}{a^2}$

Put a value

$F_1=\frac{0.1125}{0.10^2}=11.25N$

$F_2=\frac{kq_2q_3}{r^2}$

$F_2=-\frac{9\times10^9\times5\times10^{-6}\times2\times10^{-6}}{a^2}$

$F_2=-\frac{0.09}{0.01^2}=-900N$

$F=\sqrt{F_1^2+F_2^2+2F_1F_2cos\theta}$

$Cos\theta=\frac{a}{a\sqrt2}=\frac{1}{\sqrt{2}}$

Put value

$$F=\sqrt{11.25^2+(-900)^2+2\times12.25\times900\times\frac{1}{\sqrt2}}$$

$F=892.08N$

Part(b)

Magnitude of force

$|F|=892.08$ N

$\theta=tan^{-1}(\frac{900}{11.25})=-89.28°$