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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 $|\mathbf{E}|=\alpha\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$. The magnitude of constant $\alpha$ is
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The correct answer is:
$\frac{8}{3}$
The given situation is shown in the following figure
Clearly, $5 x$-components of $\mathbf{E}$ will be left to right, while $3 x$-components of $\mathbf{E}$ will be from right to left.
From the figure given below
$r^2=\frac{L^2}{4}+\frac{L^2}{2}=\frac{3 L^2}{4}$
So, $E$ due to one point charge
$|E|=\frac{K q}{r^2}=\frac{4}{3} \frac{K q}{L^2}$
Electric field at the centre of cube due to contribution of all charges is given as
$|E|_{\text {Total }}=\frac{4}{3} \frac{K q}{r^2}(5-3)=\frac{8}{3} \frac{K q}{r^2}=\frac{8}{3}\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$
Given,$|E|=\alpha\left[\frac{q}{4 \pi \varepsilon_0 L^2}\right] \quad\left[\because r^2=\frac{3 L^2}{4}\right]$
Clearly,
$\alpha=\frac{8}{3}$
Clearly, $5 x$-components of $\mathbf{E}$ will be left to right, while $3 x$-components of $\mathbf{E}$ will be from right to left.
From the figure given below
$r^2=\frac{L^2}{4}+\frac{L^2}{2}=\frac{3 L^2}{4}$
So, $E$ due to one point charge
$|E|=\frac{K q}{r^2}=\frac{4}{3} \frac{K q}{L^2}$
Electric field at the centre of cube due to contribution of all charges is given as
$|E|_{\text {Total }}=\frac{4}{3} \frac{K q}{r^2}(5-3)=\frac{8}{3} \frac{K q}{r^2}=\frac{8}{3}\left(\frac{q}{4 \pi \varepsilon_0 L^2}\right)$
Given,$|E|=\alpha\left[\frac{q}{4 \pi \varepsilon_0 L^2}\right] \quad\left[\because r^2=\frac{3 L^2}{4}\right]$
Clearly,
$\alpha=\frac{8}{3}$
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