A cube of side b has a charge q at each of it | vedclass

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(C) The length of the side of the cube is $b$.There is a charge $q$ at each of its $8$ vertices.The distance from the centre of the cube to any vertex

Vedclass · Accepted Answer

$\frac{4 q}{\sqrt{3} \pi \epsilon_{0} b}$

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(C) The length of the side of the cube is $b$.There is a charge $q$ at each of its $8$ vertices.The distance from the centre of the cube to any vertex is $r = \frac{1}{2} 	imes 	ext{diagonal of the cube}$.The diagonal of the cube is $\sqrt{b^2 + b^2 + b^2} = b\sqrt{3}$.Thus,$r = \frac{b\sqrt{3}}{2}$.The electric potential $V$ at the centre due to one charge is $V_1 = \frac{1}{4\pi\epsilon_0} \frac{q}{r}$.Since there are $8$ charges,the total potential $V = 8 	imes V_1 = 8 	imes \frac{1}{4\pi\epsilon_0} \frac{q}{(b\sqrt{3}/2)}$.$V = \frac{8q}{4\pi\epsilon_0} 	imes \frac{2}{b\sqrt{3}} = \frac{4q}{\sqrt{3}\pi\epsilon_0 b}$.The electric field at the centre due to each charge is directed away from the charge (if $q>0$) towards the opposite vertex. Since the charges are placed symmetrically,the electric field vectors from opposite vertices cancel each other out. Therefore,the net electric field at the centre is $0$.

$A$ 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.

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