Question

A cube of side \(b\) has a charge \(q\) at each of its vertices. Determine the potential and electric field due to this charge array at the centre of the cube.

Answer 1

Step 1: Draw a rough figure of given situation.
The cube of side \(b\) can be shown as follows:

Step 2: Find the distance of each vertex from the centre of the cube.
Given, length of the side of a cube =\(b\)
\(d\)= diagonal of one of the six faces of the cube
\(d^{2}= \sqrt{b^{2}+b^{2}}\)
\(d^{2}= \sqrt{2b^{2}}\)
\(d= b\sqrt{2}\)
\(l\)= lenght of the diagonal of the cube
\(l^{2}= \sqrt{b^{2}+d^{2}}\)
\(l^{2}= \sqrt{(\sqrt{2}b)^{2}}+b^{2}\)
\(l= b\sqrt{3}\)
So, the distance between the centre of the cube and one of the eight vertices.
\(r= \dfrac{l}{2}= \dfrac{b\sqrt{3}}{2}\)

Step 3: Find the electric potential at the centre of the cube.
Electric potential at the centre of the cube due to presence of the eight charges at the vertices.
\(V= \dfrac{8q}{4\pi \varepsilon _{0}r}\)
\(V= \dfrac{8q}{4\pi \varepsilon _{0}\times \dfrac{b\sqrt{3}}{2}}= \dfrac{4q}{\pi \varepsilon _{0}\times b\sqrt{3}}\)
\(V= \dfrac{4q}{\sqrt{3}\pi \varepsilon _{0}b}\)

Step 4: Find the eletric field at the centre of the cube.

As the net electric field at the centre is the vector sum of the electric field due to individual charges.

So, the electric field at the centre of the cube, due to eight charges, gets cancelled, because thde charges are distributed symmetrically with to the centre of the cube.

Hence, the electric field is zero at the centre.