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Four point charges $+ 1\mu C, + 1 \mu C, – 1 \mu C$ and $– 1 \mu C$ are placed at the corners A, B, C and D of a square of each side 0.1 m (i) Calculate electric potential at the centre O of the square (ii) If E is middle point of BC, what is work done in carrying an electron from O to E?

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In figure AB = BC = CD = DA = a = 0.1 m.
$BD=AC=\sqrt{a^2+a^2}=a \sqrt 2$
$OA=OC=OB=OD=\large\frac{1}{2} $$ a \sqrt 2=\large\frac{a}{\sqrt 2}$
Potential at O,$V_0 = \large\frac{1}{4 \pi \in_0} \times \bigg[ \large\frac{1 \times 10^{-6}}{OA}+\frac{1 \times 10^{-6}}{OB}-\frac{1 \times 10^{-6}}{OC}-\frac{1 \times 10^{-6}}{OD} \bigg]$
$\qquad =\large\frac{10^{-6}}{4 \pi \in_0} \bigg[ \large\frac{1}{a/ \sqrt {2}}+\frac{1}{a/ \sqrt {2}}-\frac{1}{a/ \sqrt {2}}-\frac{1}{a/ \sqrt {2}}\bigg]=0$
Now, $BE=CE=a/2$
Again $AE=DE= \sqrt {DC^2+CE^2}$
$\qquad= \sqrt {a^2+(a/2)^2}$
$\qquad= \sqrt {\large\frac{5a^2}{4} }$
$\qquad= \sqrt {\large\frac{a \sqrt 5}{2}}$
Potential at E,$V_E = \large\frac{1}{4 \pi \in_0} \times \bigg[ \large\frac{1 \times 10^{-6}}{AE}+\frac{1 \times 10^{-6}}{BE}-\frac{1 \times 10^{-6}}{CE}-\frac{1 \times 10^{-6}}{DE} \bigg]$
$\qquad= \large\frac{10^{-6}}{4 \pi \in_0} \bigg[ \large\frac{2}{a \sqrt 5} +\frac{2}{a} -\frac{2}{a} -\frac{2}{a \sqrt 5}\bigg]=0$
Work done in carrying an electron of charge (– e) from O to E.
$W= -e [V_E -V_o]=-e [0-0]=zero$
answered Jun 7, 2014 by meena.p

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