2.13 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.

Answers (1)

As we know,

the distance between vertices and the centre of the cube 

d=\frac{\sqrt{3}b}{2}

Where b is the side of the cube.

So Potential at the centre of the cube:

P=8*\frac{kq}{d}=8*\frac{kq}{b\sqrt{3}/2}=\frac{16kq}{b\sqrt{3}}

Hence electric potential at the centre will be

 \frac{16kq}{b\sqrt{3}}=\frac{16q}{4\pi \epsilon_0 b\sqrt{3}}=\frac{4q}{\pi \epsilon_0 b\sqrt{3}}

The electric field will be zero at the centre due to symmetry i.e. every charge lying in the opposite vertices will cancel each other's field.

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