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What will be the electric field at point P. Assume (r >> a)


$\begin{array}{1 1} (A)\;\frac{kqa^2}{r^4}\\ (B)\;\frac{6kqa^2}{r^2} \\ (C)\; \frac{6kqa^2}{r^4} \\ (D)\;\frac{6kqa^2}{r^4} \end{array}$

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$\overrightarrow {E_1}= \large\frac{q}{4 \pi \in _0 (r-a)^2 } \hat i$
$\overrightarrow {E_2}= \large\frac{-2q}{4 \pi \in _0 r^2 } \hat i$
$\overrightarrow {E_3}= \large\frac{q}{4 \pi \in _0 (r+a)^2 } \hat i$
$\overrightarrow {E_{net}}= \overrightarrow {E_1}+\overrightarrow {E_2}+\overrightarrow {E_3}$
$\qquad= \large\frac{q}{4 \pi \in _0} \bigg[\large\frac{1}{(r-a)^2}-\frac{2}{r^2}+\frac{1}{(r+a)^2}\bigg] \hat i$
$\qquad= \large\frac{q}{4 \pi \in r^2} \bigg[\bigg\{1- (\large\frac{a}{r})^{-2} \bigg\}$$-2 +\bigg\{ 1+(\large\frac{a}{r})^{-2} \bigg \}\bigg]$
So , $ r > > a$
$(1+ \alpha)^n =1+ n \alpha +\large\frac{n(n-1)}{2}$$ \alpha^2+...(\alpha << 1)$
=> $\overrightarrow { E_{net}}=\large\frac{q}{4 \pi \in _0 r^2} $$\bigg [ \bigg \{ (1- 2 (\frac{-a}{r})) +\frac{-2(-2-1)}{2} (\frac{-a}{r})^2 \bigg \}-2 + \bigg \{1- \frac{2a}{r} + \frac{-2 (-2-1)}{2} (\frac{a}{r})^2\bigg\} \bigg]$
$\qquad= \large\frac{6 a^2 q}{4 \pi \in _0 r^4}$
$\qquad= \large\frac{6 K qa^2}{r^4}$
Hence C is the correct answer.
answered Jan 13, 2014 by meena.p

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