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  • 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.

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)

best_answer

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.

Posted by

Pankaj Sanodiya

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