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, $5x$-components of $B$ will be left to right, while $3x$-components of $E$ will be from right to left.
From the figure given below

$r^2=\frac{L^2}{4}+\frac{L^2}{2}=\frac{3L^2}{4}$
So, $E$ due to one point charge
$|E|=\frac{Kq}{r^2}=\frac{4}{3}\frac{Kq}{L^2}$
Electric field at the centre of cube due to contribution of all charges is given as
$|E{|}_{\text{Total }}=\frac{4}{3}\frac{Kq}{r^2}(5-3)$
$=\frac{8}{3}\frac{Kq}{r^2}=\frac{8}{3}\left(\frac{q}{4\pi {\epsilon }_0{L}^2}\right)$
Given,
$|E|=\alpha \left[\frac{q}{4\pi {\epsilon }_0{L}^2}\right]$
$\left[\because r^2=\frac{3L^2}{4}\right]$
Clearly, $\alpha =\frac{8}{3}$