Three point charges + q each are fixed at $\large A\left( {L/\sqrt 2 ,L/\sqrt 2 ,0} \right),\,\,B\left( { - L/\sqrt 2 ,\,L/\sqrt 2 ,0} \right),C\left( {0, - L,0} \right)$ .A fourth point charge – q and mass ‘m’ is kept on z-axes at D (0,0,Z) with Z<<L and released, the minimum time taken by the fourth charge to cross the xy-plane is nearly (neglect effect of gravity)

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$\large 2\sqrt {\frac{{{\pi ^3}{ \in _0}m{L^3}}}{{3{q^2}}}}$

$\large \sqrt {\frac{{{\pi ^3}{ \in _0}m{L^3}}}{{{q^2}}}}$

$\large \sqrt {\frac{{2{\pi ^3}{ \in _0}m{L^3}}}{{3{q^2}}}}$

$\large \sqrt {\frac{{{\pi ^3}{ \in _0}m{L^3}}}{{3{q^2}}}}$

answer is D.

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Detailed Solution

$\large {F_{net}} = 3F\,\cos \theta$
$\large = - 3\frac{1}{{4\pi { \in _0}}}\frac{{{q^2}}}{{{r^2}}}\left( {\frac{z}{r}} \right)$
$\large = \frac{{ - 3{q^2}z}}{{4\pi { \in _0}{r^3}}} \approx \frac{{ - 3{q^2}z}}{{4\pi { \in _0}{L^3}}}$ $\large \left\{ \begin{gathered} \because \,r = \sqrt {{L^2} + {z^2}} \\\ \,\,\,\,r \approx L\,\,for\,z < < L \ \end{gathered} \right\}\$
$\large ma = - \frac{{3{q^2}z}}{{4\pi { \in _0}{L^3}}}$
$\large a = - {w^2}z$
$\large T = \frac{{2\pi }}{w} = 2\pi \sqrt {\frac{{4\pi { \in _0}m{L^3}}}{{3{q^2}}}}$
$\large t = \frac{T}{4} = \sqrt {\frac{{{\pi ^3}{ \in _0}m{L^3}}}{{3{q^2}}}}$