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Four point charges $\;+1 \mu C ,+1 \mu C,-1 \mu C-1 \mu C\;$ are placed at the four consecutive corners of a square of side 0.1m . What is the magnitude of electric field at E as shown in figure .

$(a)\;144\times10^{4}\;\large\frac{(5 \sqrt{5}+1)}{\sqrt{5}}\qquad(b)\;72\times10^{4}\;\large\frac{(5 \sqrt{5}+1)}{\sqrt{5}}\qquad(c)\;36\times10^{4}\;\large\frac{(5 \sqrt{5}+1)}{\sqrt{5}}\qquad(d)\;None$

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Answer : (a) $\;144\times10^{4}\;\large\frac{(5 \sqrt{5}+1)}{\sqrt{5}}$
Explanation :
E= $\large\frac{9\times10^{9}\times10^{-6}}{(\large\frac{0.1}{2})^2}$
$|E|=9\times4\times10^{3}\times10^2$
$|E|=36\times10^{5}$
$|E_{1}|=\large\frac{9\times10^{9}\times10^{-6}}{(\sqrt{5}\times\large\frac{0.1}{2})^2}$
$=\large\frac{36}{5}\times10^5$
$|E_{net}|=2|E| +2|E_{1}|+2|E_{1}|cos \theta\quad ; cos \theta=\large\frac{1}{\sqrt{5}}$
$=2 (36\times10^{5} + \large\frac{36}{5}\times10^{5} \times \large\frac{1}{\sqrt{5}})$
$=\large\frac{72\times10^5\;(5 \sqrt{5}+1)}{5 \sqrt{5}}$
$=144\times10^{4}\;\large\frac{(5 \sqrt{5}+1)}{\sqrt{5}}\;.$
answered Feb 7, 2014 by yamini.v
 

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