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Six charges are placed at the vertices of a regular hexagon as shown in the figure. The electric field on the line passing through point O and perpendicular to the plane of the figure at a distance of x ( gt gt a) from O is 1) (Qa)/(pi epsilon_0 x^3) ""2) (2Qa)/(pi epsilon_0 x^3) 3) (sqrt(3) Qa)/(pi epsilon_0 x^2) "" 4) zero. |
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Answer» <P> Solution :This is BASICALLY a problem of finding the electric field DUE to three dipoles. The DIPOLE moment of each dipole is P = Q(2a)Electric field due to each dipole will be `E = (KP)/(x^3)`. The direction of electric field due to each dipole is as shown below: `E_("net") = E + 2 E cos 60^@ = 2E` `= 2(1/(4 pi epsilon_0)) ((2Qa)/(x^3)) = (Qa)/(pi epsilon_0 x^3)`. 
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