Two charges are $+10 \mu C$ and $-10 \mu C$ are separated by $10 \mathrm{~cm}$. The magnitude of force acting on another charge $5 \mu \mathrm{C}$ placed at the midpoint of the line joining the two charges will be [Use $\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9$ in SI unit]

• Correct answer
  A

  $360 \mathrm{~N}$
• B

  $0 \mathrm{~N}$
• C

  $320 \mathrm{~N}$
• D

  $380 \mathrm{~N}$

Solution

$\begin{aligned} & \mathrm{F}=\mathrm{F}_1+\mathrm{F}_2 \\ & \Rightarrow \mathrm{F}=\frac{9 \times 10^9 \times 10 \times 5 \times 10^{-12}}{\left(5 \times 10^{-2}\right)^2} \times 2 \\ & \Rightarrow \mathrm{F}=\frac{9 \times 10 \times 5 \times 10^{-3}}{25 \times 10^{-4}} \times 2 \\ & \Rightarrow \quad \mathrm{F}=360 \mathrm{~N}\end{aligned}$

Hence, the correct answer is option: 'A'.