Question

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 center is found to be
$|\vec{ E }|=\alpha\left(\frac{q}{4 \pi \in_{0} L ^{2}}\right)$.The magnitude $=f$ constant $\alpha$ is

• A. $\frac{4}{3}$
• B. $\frac{8}{3}$
• C. $3$
• D. $1$

Answer 1

Answer: (B)

The given situation is shown in the following figure

Clearly, $5 x$-components of $B$ will be left to right, while $3 x$-components of $E$ 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] $
$\left[\because r^{2}=\frac{3 L^{2}}{4}\right]$
Clearly, $\alpha=\frac{8}{3}$