Question

Identical charges ($-q$) are placed at each corners of cube of side $b$ then electrostatic potential energy of charge ($+q$) which is placed at centre of cube will be

• A. $\frac{-4\sqrt{2}{q}^2}{\pi {\epsilon }_{0}b}$
• B. $\frac{-8\sqrt{2}{q}^2}{\pi {\epsilon }_{0}b}$
• C. $\frac{-4q^2}{\sqrt{3}\pi {\epsilon }_{0}b}$
• D. $\frac{8\sqrt{2}{q}^2}{\pi {\epsilon }_{0}b}$

Answer 1

The correct option is C $\frac{-4q^2}{\sqrt{3}\pi {\epsilon }_{0}b}$

$\text{Step 1: Distance of Central charge from charges at corners[Refer Fig.]}$

From Figure:

$O$ is center of cube

Since $YZ$ is face diagonal

$\therefore YZ=\sqrt{a^2+a^2}=\sqrt{2}a$

Since $XZ$ is body diagonal

$\therefore XZ=\sqrt{(\sqrt{2}a{)}^2+a^2}=\sqrt{3}a$

Distance between center and corner $r=\frac{XZ}{2}$

$\therefore r=\frac{\sqrt{3}a}{2}$

$\text{Step 2: Total Potential Energy}$

Potential energy of two charges $U=\frac{K{q}_1{q}_2}{r}$

Here $8(-q)$ charges are placed at corners of the cube.

And all these charges are at same distance from central charge $q$.

Since, Potential energy is scalar quantity, so it will be added due all the corner charges.

Therefore, $U_{Total}=8\times U_1$

$\Rightarrow U_T=8\times \frac{K{q}_1{q}_2}{r}$

$=\frac{8K(+q)(-q)}{\left(\frac{\sqrt{3}a}{2}\right)}$ $=\frac{-16K{q}^2}{\sqrt{3}a}$

So, $U_T=\frac{-4q^2}{\sqrt{3}\pi {\in }_{0}a}$

Hence Option ${}^{\prime }{C}^{\prime }$ is correct.