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.

Given., length of side of cube = b

and charge at each side of its vertices = q

As we know,

the length of diagonal of the cube side, d = √(b2 + b2) = √(2b2)= b√2

Let l be the length of diagonal of cube. Thus,

l = √(d2 + b2)

= √(2b2+b2)

= √3b2

= b√3

Distance between center of cube and vertices, r = l/2

r = (b√3)/2

The electric potential (V) at the centre of the cube is due to the presence of 8 charges at the vertices.

So,

[latex]\begin{aligned} V &=\frac{8 q}{4 \pi \epsilon_{0}} \\ &=\frac{8 q}{4 \pi \in_{0}\left(b \frac{\sqrt{3}}{2}\right)} \\ &=\frac{4 q}{\sqrt{3} \pi \in_{0} b} \end{aligned}[/latex]

The electric field intensity at the center due to all the eight charges is zero because the fields due to the presence of charges at the vertices cancel in pairs.

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