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A charge $+q$ is situated at a distance $d$ away from both the sides of a grounded conducting $L$ shaped sheet as shown in the figure.The force acting on the charge $+q$ is
towards $O$, magnitude $\frac{q^2}{32 \pi \varepsilon_0 d^2}(2 \sqrt{2}+1)$
away from $O$, magnitude $\frac{q^2}{32 \pi \varepsilon_0 d^2}(2 \sqrt{2}+1)$
towards $O$, magnitude $\frac{q^2}{32 \pi \varepsilon_0 d^2}(2 \sqrt{2}-1)$
away from $O$, magnitude $\frac{q^2}{32 \pi \varepsilon_0 d^2}(2 \sqrt{2}-1)$
Solution
(c)
The given diagram can be shown by symmetry of four charges as,
Let $O$ be the centre and $2 d$ be the distance between charges.
Forces between $-q$ and $+q$ is
$F_1=\frac{K q^2}{4 d^2}=F_2 \text { (attractive) }$
and force between $+q$ and $+q$ is
$F_3=\frac{K q^2}{(2 \sqrt{2} d)^2}=\frac{K q^2}{8 d^2} \text { (repulsive) }$
So, net force acting on the charge is
$F_{\text {net }}=F_{12}-F_3=\sqrt{2} F_1-F_3$
$\left(\because F_{12}\right.=\sqrt{\left.F_1^2+F_1^2=\sqrt{2} F_1\right)}$
$=\sqrt{2} K q^2-\frac{K q^2}{8 d^2}$
$=\frac{q^2(2 \sqrt{2}-1)}{32 \pi \varepsilon_0 d^2}, \text { towards } O$
