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# Dipole in uniform electric eld Consider a dipole in a uniform electric eld as shown in the gure below

Let the lengh of the dipole be such that r r s s r r and the forces on and be and q 8722q qE 8722qE The total force on the dipole in the 64257eld is zero 0 but there is a torque about the midpoint of the dipole which is r r s s q s p 5 The direc

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## Dipole in uniform electric eld Consider a dipole in a uniform electric eld as shown in the gure below

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## Presentation on theme: "Dipole in uniform electric eld Consider a dipole in a uniform electric eld as shown in the gure below"— Presentation transcript:

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Dipole in uniform electric ﬁeld : Consider a dipole in a uniform electric ﬁeld as shown in the ﬁgure below. Let the lengh of the dipole be , such that ~r ~r ~s/ , ~s ~r ~r and the forces on + and be and +q −q qE −qE The total force on the dipole in the ﬁeld is zero, = 0 but there is a torque about the midpoint of the dipole, which is ~r ~r ~s ~s q ~s ~p (5) The direction of is such as to line up ~p parallel to Dipole in non-uniform electric ﬁeld : This time around the total force = 0, there is a net force on dipole in addition to

torque. Let the el ectric ﬁeld close to + is while at . The total force then is, qd We make use of the relation d to re-write as = ( d~s . Hence, we have qd d~s and ~p qd~s and therefore, = ( ~p (6) The torque in non-uniform ﬁeld, using the notation ~r ~r ~r is, ~r ~r ~r ~r ~r ~r ~r ~s ~r ~p ~r ~p (7) Energy of a dipole in electric ﬁeld : We calculate the energy of a diple in a electric ﬁeld by bringing the dipole from inﬁnity and placing it such that charge is at ~r . Work done is ( )), q V ) + q V Now, since , we can Taylor expand ) as, ) = ) + ~s ) +

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Putting back the above expression in , we get q V ) + q V ) + q ~s ~p ~p (8) Work done to rotate a dipole : Rotating a dipole kept in a uniform electric ﬁeld by an angle requires work, Nd p E sin θ d p E (1 cos ) (9) Field due to di-polar object : Consider a dipole consists of two equal and opposite charges ( ) separated by a distance . The object itself may be a dielectric or a molecule, but it is placed in vacuum, that is why we still be using . Let and be the distances of the ﬁeld point from the charges and dipole center, where −q +q r+ r The

potential at point P is, ) = and from law of cosines, + ( s/ 2) rs cos cos Since we are interested in the region , the ) term is negligibly small and binomial expansion yields, cos cos Using the above the expression, we get cos and the potential for dipole at point P is, therefore, given b ) = qs cos ~p (10)