Question

A cube of side 'a' has a charge q placed at each of its eight corners. The potential at the centre of the cube due to all the charges is:
A) $\dfrac{16q}{4\pi {{\varepsilon }_{0}}a\sqrt{3}}$
B) $\dfrac{16q}{4\pi {{\varepsilon }_{0}}a}$
C) $\dfrac{q}{4\pi {{\varepsilon }_{0}}a}$
D) $\dfrac{q}{4\pi {{\varepsilon }_{0}}a\sqrt{3}}$

Answer 1

Here, we calculate potential at centre due to each charge. The distance between each charge and centre is equal to half of the diagonal. The total potential due to 8 charges at the corner will give potential at the centre.

Complete answer:
A diagram can be illustrated as follows:

Let the cube of side be 'a’
So the length of each diagonal is given by $D=\sqrt{{{a}^{2}}+{{a}^{2}}+{{a}^{2}}}$
$=\sqrt{3{{a}^{2}}}=\sqrt{3}a$
Distance of each corner from the Centre O is half of its diagonal$r=\dfrac{\sqrt{3}a}{2}$
Potential at O due to charge q at each centre is given by
$V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}.\dfrac{a}{r}=\dfrac{1}{4\pi {{\varepsilon }_{0}}}.\dfrac{q}{\dfrac{\sqrt{3}a}{2}}$
$=\dfrac{2q}{4\pi {{\varepsilon }_{0}}\sqrt{3}a}$
Therefore, net potential at O due to all the 8 changes at the corners of the cubes $V=8\times \dfrac{2q}{4\pi {{\varepsilon }_{0}}\sqrt{3}a}$
$V=\dfrac{16q}{4\pi {{\varepsilon }_{0}}\sqrt{3}a}$
The electric field at O due to charge at all the corners of the cube is zero, since the electric field due to charges at opposite 8 corners are equal and opposite.

Note:
As we know that the cube of side a and diagonal is $D=\sqrt{3}a$. So, Be careful during the calculation of distance of each corner from the Centre is half of its diagonal i.e. $r=\dfrac{\sqrt{3}a}{2}$ and the potential $V=\dfrac{1}{4\pi {{\varepsilon }_{0}}}\dfrac{2q}{\sqrt{3}a}$ For eight corners we multiplied it by 8 we get the results.