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(B) Let the centre of the cube be the origin $(0,0,0)$. The vertices of the cube are at $(\pm L/2, \pm L/2, \pm L/2)$.If all eight vertices had a char

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$\frac{8}{3}$

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(B) Let the centre of the cube be the origin $(0,0,0)$. The vertices of the cube are at $(\pm L/2, \pm L/2, \pm L/2)$.If all eight vertices had a charge $+q$,the electric field at the centre would be zero due to symmetry.Let the charge at one vertex (say $(-L/2, -L/2, -L/2)$) be $-q$ instead of $+q$. Let $\vec{E}_{total}$ be the electric field at the centre with the given configuration.Let $\vec{E}_{all}$ be the electric field if all eight vertices had charge $+q$,so $\vec{E}_{all} = 0$.Let $\vec{E}_{vertex}$ be the electric field at the centre due to a charge $+q$ at the vertex $(-L/2, -L/2, -L/2)$.Then,$\vec{E}_{total} = \vec{E}_{all} - \vec{E}_{vertex} + \vec{E}_{(-q)} = 0 - \vec{E}_{vertex} - \vec{E}_{vertex} = -2\vec{E}_{vertex}$.The distance $r$ from any vertex to the centre is $r = \sqrt{(L/2)^2 + (L/2)^2 + (L/2)^2} = \sqrt{3L^2/4} = \frac{\sqrt{3}L}{2}$.The magnitude of the electric field due to one charge $q$ at the centre is $|E_{vertex}| = \frac{1}{4\pi\varepsilon_0} \frac{q}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q}{3L^2/4} = \frac{4}{3} \left(\frac{q}{4\pi\varepsilon_0 L^2}ight)$.Since $\vec{E}_{total} = -2\vec{E}_{vertex}$,the magnitude is $|E_{total}| = 2 |E_{vertex}| = 2 	imes \frac{4}{3} \left(\frac{q}{4\pi\varepsilon_0 L^2}ight) = \frac{8}{3} \left(\frac{q}{4\pi\varepsilon_0 L^2}ight)$.Comparing this with $|E| = \alpha \left(\frac{q}{4\pi\varepsilon_0 L^2}ight)$,we get $\alpha = \frac{8}{3}$.

$A$ cube of side $L$ has point charges $+q$ located at its seven vertices and $-q$ at remaining one vertex. The electric field at its centre is found to be $|E|=\alpha\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$. The magnitude of constant $\alpha$ is

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