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Three equal charges, each having a magnitude of $2.0 \times 10^{-6} \mathrm{C}$, are placed at the three corners of a right angled triangle of sides $3 \mathrm{~cm}, 4 \mathrm{~cm}$ and 5 cm . The force (in magnitude) on the charge at the right angled corner is
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The correct answer is:
45.9 N
Consider the diagram
The situation is shown in figure.
The force on $A$ due to $B$ is
$F_1=\frac{\left(2.0 \times 10^{-6}\right)\left(2.0 \times 10^{-6}\right)}{4 \pi \varepsilon_0(4 \mathrm{~cm})^2}$
$\begin{aligned}=9 \times 10^9 \times 4 \times 10^{-12} & \times \frac{1}{16 \times 10^{-4}} \\ & =22.5 \mathrm{~N}\end{aligned}$
As the force due to $C$ on $A$ is 40 N . Thus, net force (electric)
$\begin{aligned} F & =\sqrt{F_1^2+F_1^2} \\ & =\sqrt{(22.5)^2+(40)^2}=45.9 \mathrm{~N}\end{aligned}$
The situation is shown in figure.
The force on $A$ due to $B$ is
$F_1=\frac{\left(2.0 \times 10^{-6}\right)\left(2.0 \times 10^{-6}\right)}{4 \pi \varepsilon_0(4 \mathrm{~cm})^2}$
$\begin{aligned}=9 \times 10^9 \times 4 \times 10^{-12} & \times \frac{1}{16 \times 10^{-4}} \\ & =22.5 \mathrm{~N}\end{aligned}$
As the force due to $C$ on $A$ is 40 N . Thus, net force (electric)
$\begin{aligned} F & =\sqrt{F_1^2+F_1^2} \\ & =\sqrt{(22.5)^2+(40)^2}=45.9 \mathrm{~N}\end{aligned}$
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