Six charges are placed at the certices 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 as a function of distance x from point O is (x>>a)
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we know, electric field due to dipole P on its axis is given by
E = KP/x³
where K = 1/4π∈₀ , x is the separation between midpoint of dipole to observation point. Here side length is a , means distance between two opposite nature charge is 2a . So, dipole , P = 2Qa
∴E = Qa/2π∈₀x³ ------(1)
Now, see figure sincerely,
Electric field due to charge Placed at E and electric field due to charge place at B is in same direction.
So, electric field along OB = E
Similarly, electric field along OC = E
electric field along OD = E
See 3rd attachment, for finding Enet
|Enet| = 2E [ because resultant of two electric fields in which Angle between them is 120° is E along OC . So, Enet = resultant of OB and OD + electric field of OC = E + E = 2E ]
so, Enet = 2E = 2Qa/2π∈₀x³ = Qa/π∈₀x³
Hence, answer is Qa/π∈₀x³
E = KP/x³
where K = 1/4π∈₀ , x is the separation between midpoint of dipole to observation point. Here side length is a , means distance between two opposite nature charge is 2a . So, dipole , P = 2Qa
∴E = Qa/2π∈₀x³ ------(1)
Now, see figure sincerely,
Electric field due to charge Placed at E and electric field due to charge place at B is in same direction.
So, electric field along OB = E
Similarly, electric field along OC = E
electric field along OD = E
See 3rd attachment, for finding Enet
|Enet| = 2E [ because resultant of two electric fields in which Angle between them is 120° is E along OC . So, Enet = resultant of OB and OD + electric field of OC = E + E = 2E ]
so, Enet = 2E = 2Qa/2π∈₀x³ = Qa/π∈₀x³
Hence, answer is Qa/π∈₀x³
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