Question

# A thick conductor having inner radius a and outer radius b has a small passage as shown in the figure. A charge +Q is placed at the centre. The work done by external agent in taking the charge from centre to infinity is

A
Q24πε0aQ24πε0b+Q24πε0(ab)
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B
Q28πε0b
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C
Q28πε0a
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D
Q28πε0aQ28πε0b
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Solution

## The correct option is D Q28πε0a−Q28πε0bThe induced charge will be as shown in the figure, Initial electrostatic energy of the given system at the center of the shell is Ui=(Uself)1+(Uself)2+[Ineraction energy].....(1) Where, (Uself)1=Electrostatic potential energy at the centre of the shell due to charge induced at inner surface=Q28πε0a (Uself)2=Electrostatic potential energy at the centre of the shell due to charge induced at outer surface=Q28πε0b Interaction energy E= Electrostatic potential energy due to induced charges and the centre charge=−Q24πε0a−Q24πε0b+Q24πε0b Substituting the values in equation 1 we get, Ui=Q28πε0a+Q28πε0b−Q24πε0a−Q24πε0b+Q24πε0b ⇒Ui=Q28πε0b−Q28πε0a When the charge is taken to infinite the electrostatic potential energy of the system will be zero.i.e., Uf=0 Work done by external agent in taking the charge from centre to infinity is Wext=Uf−UiSubstituting the values, we have Wext=0−(Q28πε0b−Q28πε0a) ∴Wext=−Q28πε0b+Q28πε0a Hence, option (a) is correct option.

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