A cube of side 'b' has charge 'q' at each of its vertices. The electric field at the centre of the cube will be

zero

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\frac{32 q}{b^{2}}

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\frac{q}{2 b^{2}}

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\frac{q}{b^{2}}

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Solution

The correct option is A zero
The electric field is a vector quantity.
At the center of the cube, at which the electric fields are equal in magnitude and opposite in directions, due to charges at opposite vertices of the cube, the resultant Electric field at the center would be zero.

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A cube of side 'b' has charge 'q' at each of its vertices. The electric field at the centre of the cube will be

The correct option is A zeroThe electric field is a vector quantity.At the center of the cube, at which the electric fields are equal in magnitude and opposite in directions, due to charges at opposite vertices of the cube, the resultant Electric field at the center would be zero.

The correct option is A zero
The electric field is a vector quantity.
At the center of the cube, at which the electric fields are equal in magnitude and opposite in directions, due to charges at opposite vertices of the cube, the resultant Electric field at the center would be zero.