A charge +q is situated at a distance d away | vedclass

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(C) Using the method of images for a grounded conducting corner,we place three image charges to satisfy the boundary conditions: $-q$ at $(-d, 0)$,$-q

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towards $O$,magnitude $\frac{q^2}{32 \pi \varepsilon_0 d^2}(2 \sqrt{2}-1)$

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(C) Using the method of images for a grounded conducting corner,we place three image charges to satisfy the boundary conditions: $-q$ at $(-d, 0)$,$-q$ at $(0, -d)$,and $+q$ at $(-d, -d)$.The force on the charge $+q$ at $(d, d)$ due to the image charges is:$1$. Force due to $-q$ at $(-d, 0)$: $F_1 = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(2d)^2 + 0^2} = \frac{q^2}{16 \pi \varepsilon_0 d^2}$ (attractive,towards the $x$-axis).$2$. Force due to $-q$ at $(0, -d)$: $F_2 = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{0^2 + (2d)^2} = \frac{q^2}{16 \pi \varepsilon_0 d^2}$ (attractive,towards the $y$-axis).$3$. Force due to $+q$ at $(-d, -d)$: $F_3 = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(2d)^2 + (2d)^2} = \frac{q^2}{32 \pi \varepsilon_0 d^2}$ (repulsive,away from the origin).The resultant attractive force $F_{12}$ from the two $-q$ charges is $\sqrt{F_1^2 + F_2^2} = \sqrt{2} F_1 = \frac{\sqrt{2} q^2}{16 \pi \varepsilon_0 d^2}$,directed towards the origin $O$.The net force $F_{	ext{net}}$ is $F_{12} - F_3 = \frac{\sqrt{2} q^2}{16 \pi \varepsilon_0 d^2} - \frac{q^2}{32 \pi \varepsilon_0 d^2} = \frac{q^2}{32 \pi \varepsilon_0 d^2} (2\sqrt{2} - 1)$,directed towards $O$.

$A$ 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:

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