Question Solved1 Answer Two dipoles (blue and orange) are aligned down their middles asshown in the diagram. The dipoles’ charges and separations are alsoshown in the diagram. The distance between the centers of the twodipoles is d, which is much bigger than the dipoleseparations. Calculate the net electric field at a location exactlyhalfway between the centers of the two dipoles. 3,9 + - d + 2s, a
Two dipoles (blue and orange) are aligned down their middles as
shown in the diagram. The dipoles’ charges and separations are also
shown in the diagram. The distance between the centers of the two
dipoles is d, which is much bigger than the dipole
separations. Calculate the net electric field at a location exactly
halfway between the centers of the two dipoles.
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We have to trind net Electric fied at point e.Soln:We know that electric field due to charge ' q ' at distance r is given by (kq)/(r^(2))i). if q is +ve. electric field is away from charge.ii) if q is -ve, Direction of Electric field toward the charge.E_(PA)&E_(PB) is Electric field At foint P due to A _|_ B.Magnitude of E_(PA)= Magnitude of E_(PB)=Erarr Electric field in Horizontal direction at point ' P ' due to Blue dipale is 2E cos theta.EDMINOTE P PRO MAXQuad cameraElectric field in vertical dprection at point point ' p ' due to Blue dipole is zero.Enet at point ' rho ' due to Blue dipole{:[=2E cos theta+0longrightarrow],[=2xx(Kq)/([sqrt(((s)/(2))^(2)+((d)/(2))^(2))]^(2))(S)/(2*sqrt(((s)/(2))^(2)+((d)/(2))^(2)))]:}=(2kqs)/(2[((s)/(2))^(2)+((9)/(2))^(2)]^(3//2))11=(kqs)/([((5)/(2))^(2)+((d)/(2))^(2)]^(3//2))=(kqs)/(1//2^(3)(s^(2)+d^(2))^(3//2))In the Question, 9t is given thatd>>sSo, we can write s^(2)+d^(2)~~d^(2)=(" kqs ")/(1//2^(3)d^(3))=(2^(3)kqs)/(d^(3))l ... See the full answer