IIT JEE past question Electrostatics 1

JEE Mains 2015 Papers › Electricity and magnetism questions › IIT JEE past question Electrostatics 1

A charge +Q is fixed at the origin. Far away from it an electric dipole of dipole moment P pointing away from the origin is released from rest. The length of the dipole is negligible.

Find
i) the kintetic energy of the dipole at a distance d from the origin
ii) the magnitude of the force n charge Q due to the dipole when it is at this position.

#Physics #Electricity and magnetism

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6 Answers

1

voldy ·2009-03-29 07:12:53

hey

for first one
i) the total KE =sum of rotational and translational kinetic energies .
this is bcos in non-uniform field dipole both rotates and translates.

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21

tapanmast Vora ·2009-03-31 03:07:34

ne1?

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1

Samarth Kashyap ·2009-03-31 03:32:41

is my answer correct?????

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21

tapanmast Vora ·2009-03-31 03:34:59

"The length of the dipole is negligible."

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1

skygirl ·2009-03-31 09:05:44

1.) KE = kQp/d²

2) magnitude of force on charge Q = k2QP/d³

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1

skygirl ·2009-03-31 09:16:47

electric field due to Q on the dipole = -k 2Qx/d³ [derive n see]

potential = -E.d = k.2Qx/d² = V

W = Vq = 2Qkp/d²

U = -W = -2Qkp/d²

kinetic energy = -1/2 .U = Qkp/d²

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