Question

A cube of side '$a$' has a charge $q$ placed at each of its eight corneres. The potential at the centre of the cube due to all the charges is :

• A. $\frac{16q}{4\pi {\epsilon }_{0}a\sqrt{3}}$
• B. $\frac{16q}{4\pi {\epsilon }_{0}a}$
• C. $\frac{q}{4\pi {\epsilon }_{0}a}$
• D. $\frac{q}{4\pi {\epsilon }_{0}a\sqrt{3}}$

Answer 1

The correct option is A $\frac{16q}{4\pi {\epsilon }_{0}a\sqrt{3}}$

Side of the cube $=a$

Length of each diagonal $=\sqrt{a^2+a^2+a^2}=\sqrt{3}a$

Distance of each corner from centre $O=$ half the diagonal $=\frac{\sqrt{3}a}{2}$

Potential at $O$ due to charge at each corner $=\frac{1}{4\pi {\epsilon }_o}\frac{q}{\sqrt{3}\frac{a}{2}}$ = $\frac{1}{4\pi {\epsilon }_o}\frac{2q}{\sqrt{3}a}$

Therefore, net potential at $O$ due to all the $8$ charges at the corners of the cube,$V=8\times \frac{1}{4\pi {\epsilon }_o}\frac{2q}{\sqrt{3}a}=\frac{1}{4\pi {\epsilon }_o}\frac{16q}{a\sqrt{3}}$

The electric field at $O$ due to charges at all the corners of cube is zero, since, the electric fields due to charges at opposite corners such as $A$ and $H$, $G$ and $E$, $F$ and $C$ are equal and opposite.