Question

Four equal charges of 2.0 × 10⁻⁶ C each are fixed at the four corners of a square of side 5 cm. Find the Coulomb's force experienced by one of the charges due to the other three.

Answer 1

Given,
Magnitude of the charges, $q=2\times {10}^{-6}\mathrm{C}$
Side of the square, $a=5\mathrm{cm}=0.05\mathrm{m}$
By Coulomb's Law, force,
$F=\frac{1}{4\mathrm{\pi }{\epsilon }_0}\frac{q_1{q}_2}{r^2}$

So, force on the charge at A due to the charge at B,
$$\begin{gathered} {\overrightarrow{F}}_{\mathrm{BA}}=\frac{9\times {10}^9\times {\left(2\times {10}^{-6}\right)}^2}{{\left(0.05\right)}^2} \\ =\frac{9\times {10}^9\times 4\times {10}^{-12}}{25\times {10}^{-4}} \\ =14.4\mathrm{N} \end{gathered}$$
Force on the charge at A due to the charge at C,
$$\begin{gathered} {\overrightarrow{F}}_{\mathrm{CA}}=\frac{9\times {10}^9\times {\left(2\times {10}^{-6}\right)}^2}{{\left(\sqrt{2}\times 0.05\right)}^2} \\ =\frac{9\times {10}^9\times 4\times {10}^{-12}}{25\times 2\times {10}^{-4}} \\ =7.2\mathrm{N} \end{gathered}$$
Force on the charge at A due to the charge at D,
${\overrightarrow{F}}_{\mathrm{DA}}={\overrightarrow{F}}_{\mathrm{BA}}$

The resultant force at A, F'= ${\overrightarrow{F}}_{\mathrm{BA}}+{\overrightarrow{\mathit{F}}}_{\mathrm{CA}}+{\overrightarrow{\mathit{F}}}_{\mathrm{DA}}$
The resultant force of ${\overrightarrow{F}}_{\mathrm{DA}}\mathrm{and}{\overrightarrow{F}}_{\mathrm{BA}}$ will be $\sqrt{2}{F}_{\mathrm{BA}}$ in the direction of ${\overrightarrow{F}}_{\mathrm{CA}}$. Hence, the resultant force,
$$\begin{gathered} F\text{'}=14.4\sqrt{2}+7.2 \\ =27.56\mathrm{N} \end{gathered}$$