Question 780: Electrostats & Capacitance

Question

The electric charges ' $+2 q$ ', ' $+2 q$ ', ' $-2 q$ ' and ' $-2 q$ ' are placed at the corners of square of side ' 2 L ' as shown in figure. The electric potential at point 'A', midway between the two charges ' $+2 q$ ' and ' $+2 q$ ' is
( $\varepsilon_0=$ permittivity of free space)

Accepted Answer

$$ \begin{aligned} &\begin{array}{ll}  & A C^2=(2 L)^2+L^2=4 L^2+L^2=5 L^2 \ 	herefore \quad & A C=A D=\sqrt{5} L \ & A B=A E=L \end{array}\ &	herefore \quad 	ext { Potential at point } \mathrm{A} 	ext { due to the four charges }\ &\begin{aligned} V & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{2 q}{L}+\frac{2 q}{L}-\frac{2 q}{\sqrt{5} L}-\frac{2 q}{\sqrt{5} L}ight) \ & =\frac{1}{4 \pi \varepsilon_0}\left(\frac{4 q}{L}-\frac{4 q}{\sqrt{5} L}ight) \ & =\frac{1}{4 \pi \varepsilon_0} \cdot \frac{4 q}{L}\left(1-\frac{1}{\sqrt{5}}ight) \ & =\frac{q}{\pi \varepsilon_0 L}\left(1-\frac{1}{\sqrt{5}}ight) \end{aligned} \end{aligned} $$

Answer

$\quad \frac{1}{4 \pi \varepsilon_0}\left(\frac{2 q}{L}\right)\left[1+\frac{1}{\sqrt{5}}\right]$

Answer

$\frac{\mathrm{q}}{\pi \varepsilon_0 \mathrm{~L}}\left[1-\frac{1}{\sqrt{5}}\right]$

Answer

$\frac{\mathrm{q}}{\pi \varepsilon_0 \mathrm{~L}}\left[1+\frac{1}{\sqrt{5}}\right]$

Answer

$\frac{1}{4 \pi \varepsilon_0}\left(\frac{2 q}{L}\right)\left[1-\frac{1}{\sqrt{5}}\right]$

Electrostats & Capacitance

The electric charges ' +2 q ', ' +2 q ', ' -2 q ' and ' -2 q ' are placed at the corners of square of side ' 2 L ' as shown in figure. The electric potential at point 'A', midway between the two charges ' +2 q ' and ' +2 q ' is ( \varepsilon_0= permittivity of free space)

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