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

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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

$V_{P}=\large\frac{2\;k\;q}{a}-\large\frac{2\;k\;q}{a(1+\large\frac{b^2}{a^2})^{\large\frac{1}{2}}}$

$=\large\frac{2\;k\;q}{a}-\large\frac{2\;k\;q}{a}\;[1-\large\frac{1}{2} \large\frac{b^2}{a^2}]$

$=\large\frac{k\;q\;b^2}{a^3}\;.$

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