Question

A cube of side x has a charge q at each of its vertices. The potential due to this charge array at the centre of the cube is.

• A. $\frac{4q}{3\pi\varepsilon_{0}x}$
• B. $\frac{4q}{\sqrt{3}\pi\varepsilon_{0}x}$
• C. $\frac{3q}{4\pi\varepsilon_{0}x}$
• D. $\frac{2q}{\sqrt{3}\pi\varepsilon_{0}x}$

Answer 1

Answer: (B)

$\frac{4q}{\sqrt{3}\pi\varepsilon_{0}x}$

Answer 2

: The length of diagonal of the cube of each side

x is $\sqrt{3x^{2}} = x\sqrt{3}$

$\therefore$ Distance between centre of cube and each vertex,

$r = \frac{x\sqrt{3}}{2}$

Now, potential, $V = \frac{1}{4\pi\varepsilon_{o}}\frac{q}{r}$

Since cube has 8 vertices and 8 charges each of value q are present there

$\therefore V = \frac{1}{4\pi\varepsilon_{o}}\frac{8q}{\frac{x\sqrt{3}}{2}} = \frac{4q}{\sqrt{3}\pi\varepsilon_{o}x}$