Three equal charges, each having a magnitude of $2 \cdot 0 imes 10^{-6} \mathrm{C}$, are placed at

Question

Three equal charges, each having a magnitude of $2 \cdot 0 	imes 10^{-6} \mathrm{C}$, are placed at the three corners of a rightangled triangle of sides $3 \mathrm{~cm}, 4 \mathrm{~cm}$ and 5 cm . The magnitude of the force on the charge at the right-angle corner.

Professor PC Thomas Classes · Answer

The situation is shown in figure. The force on $A$ due to $B$ is
$
\begin{aligned}
F_1 & =\frac{\left(2.0 	imes 10^{-6} \mathrm{C}ight)\left(2.0 	imes 10^{-6} \mathrm{C}ight)}{4 \pi \varepsilon_0(4 \mathrm{~cm})^2} \
& =9 	imes 10^9 \mathrm{~N} \mathrm{~m}^2 \mathrm{C}^{-2} 	imes 4 	imes 10^{-12} \mathrm{C}^2 	imes \frac{1}{16 	imes 10^{-4} \mathrm{~m}^2} \
& =22.5 \mathrm{~N}
\end{aligned}
$
This force acts along $B A$. Similarly, the force on $A$ due to $C$ is $F_2=40 \mathrm{~N}$ in the direction of $C A$. Thus, the net electric force on $A$ is
$
\begin{aligned}
F & =\sqrt{F_1^2+F_2^2} \
& =\sqrt{(22.5 \mathrm{~N})^2+(40 \mathrm{~N})^2}=45.9 \mathrm{~N} .
\end{aligned}
$
This resultant makes an angle $	heta$ with $B A$ where
$
	an 	heta=\frac{40}{22 \cdot 5}=\frac{16}{9} .
$