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The electrostatic potential due to the charge configuration at point P as shown in figure is $\;(b < < a)$

$(a)\;\large\frac{2\;k\;q}{a}\qquad(b)\;\large\frac{2\;k\;q\;b^2}{a^3}\qquad(c)\;\large\frac{k\;q\;b^2}{a^3}\qquad(d)\;zero$

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Answer : (c) $\;\large\frac{k\;q\;b^2}{a^3}$
Explanation :
$V_{P}=\large\frac{2\;k\;q}{a}-\large\frac{2\;k\;q}{\sqrt{a^2+b^2}}$
Using binomial expansion
$V_{P}=\large\frac{2\;k\;q}{a}-\large\frac{2\;k\;q}{a(1+\large\frac{b^2}{a^2})^{\large\frac{1}{2}}}$
$=\large\frac{2\;k\;q}{a}-\large\frac{2\;k\;q}{a}\;[1-\large\frac{1}{2} \large\frac{b^2}{a^2}]$
$=\large\frac{k\;q\;b^2}{a^3}\;.$
answered Feb 11, 2014 by yamini.v
 

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