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Q.
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
Electrostatic Potential and Capacitance
Solution:
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, $V =\frac{1}{4\pi\varepsilon_{0}} \frac{q}{r}$
Since cube has $8$ vertices and $8$ charges each of value $q$ are present there
$\therefore V=\frac{1}{4\pi\varepsilon_{0}} \frac{8q}{\frac{x\sqrt{3}}{2}}
=\frac{4q}{\sqrt{3}\pi\varepsilon_{0} x}$