A cube of side 'a' has a charge Q at each of its vertices, what is the potential due to this charge array at the centre of the cube?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\frac{4Q}{\sqrt{3}πε_0a}$
To calculate the potential at the center of the cube due to charges Q at each of its vertices,
$V=\frac{KQ}{r}$
$r=\frac{Body\,Diagonal}{2}=\sqrt{\frac{a^2+2a^2}{2}}=\frac{\sqrt{3}a}{2}$
$∴V_{total}=8×\frac{KQ}{\frac{\sqrt{3}a}{2}}=\frac{16KQ}{\sqrt{3}a}=\frac{4Q}{\sqrt{3}πε_0a}$