QJEEAdv08Phca36: JEE Advanced 2008 Physics - Class 12 JEEAdv08Phca36 Solution: Consider a system of three charges and placed at points A, B…

Electric field at point O is vector sum of electric field due to all three point charges individually.

$\overrightarrow E = \overrightarrow {{E_A}} + \overrightarrow {{E_B}} + \overrightarrow {{E_C}}$ and $\overrightarrow {{E_A}} = - \overrightarrow {{E_B}}$

So, $\overrightarrow E = \overrightarrow {{E_C}} = {{2q/3} \over {4\pi {\varepsilon _0}.{R^2}}}$ along $\overrightarrow {OC}$

$= {q \over {6\pi \varepsilon _0^2.{R^2}}}$ along $\overrightarrow {OC}$

$= {{ + q'} \over {6\pi \varepsilon _0^2.{R^2}}}\widehat i$

$= {{ - q'} \over {6\pi \varepsilon _0.{R^2}}}\widehat j$

From geometry, we can find that $\angle$ ABC = 30 $^\circ$ and $\angle$ ACB = 90 $^\circ$

So, AB = 2R, AC = R, BC = $\sqrt3$ R

Total potential energy of the system is,

$U = {1 \over {4\pi {\varepsilon _0}}}\left[ {{{q/3 \times q/3} \over {2R}} - {{q/3 \times 2q/3} \over R} - {{q/3 \times 2q/3} \over {\sqrt 3 R}}} \right] \ne 0$

$F = {{q/3 \times q/3} \over {4\pi {\varepsilon _0} \times {{(\sqrt 3 R)}^2}}} = {{{q^2}} \over {54\pi {\varepsilon _0}{R^2}}}$

Potential at point O is,

$V = {{q/3 \times q/3 - 2q/3} \over {4\pi {\varepsilon _0}R}} = 0$