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Find electric field at point $(a,a,a) $ due to three infinitely long lines of charge density $\lambda$ along $x,y $ and $ z$ axis

$(A)\; \frac{\lambda} {3 \pi \in _0 a} (\hat i +\hat j + \hat k) \\ (B)\;\frac{\lambda} {2 \sqrt 2 \pi \in _0 a} (\hat i +\hat j + \hat k) \\ (C)\; \frac{\lambda} {2 \pi \in _0 a} (\hat i +\hat j + \hat k) \\ (D)\;\frac{\sqrt 2\lambda} { \pi \in _0 a} (\hat i +\hat j + \hat k) $

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Due to wire along z-axis,
$\overrightarrow {E_3}=\large\frac{\lambda}{2 \pi \in _0 (a \sqrt 2)} $$\quad ( \cos 45^{\circ} \hat i + \cos 45^{\circ} \hat j)$
=> $\overrightarrow {E_3} =\large\frac{\lambda}{4 \pi \in _0 a} $$(\hat i +\hat j)$
$E_1=\large\frac{\lambda}{4 \pi \in_0 a}$$(\hat j+\hat k)$
=> $\overrightarrow {E_2} =\large\frac{\lambda}{4 \pi \in _0 a} $$(\hat i +\hat k)$
=> $\overrightarrow {E}_{resultant } =\overrightarrow {E_1}+\overrightarrow {E_2}+\overrightarrow {E_3}$
$\large\frac{\lambda} {2 \sqrt 2 \pi \in _0 a} (\hat i +\hat j + \hat k)$
Hence B is the correct answer.
answered Jan 13, 2014 by meena.p
edited Aug 22, 2014 by meena.p

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