1answer is D
A point charge q = 1μC is brought from infinity (slowly so that heat developed in the shell is negligible) and is placed at the centre of a conducting neutral spherical shell of inner radius a = 1m and outer radius b = 2m, then work done by external agent is:
(A)0
(B)0.25 mJ
(C)2.25mJ
(D)-2.25mJ
-
UP 0 DOWN 0 1 8
8 Answers
1Yes,you are getting the right answer.But would you like to show me the solution?
1It is the total energy of the system (ie, Self Energy + Interaction Energy)
1yeah its some fancy concept acc to me my sol is is simple
first the potenital at the outer surface is q/4pie epsilon 2
now let us take the inner cavity the electirc field due to the conductor will be q/4pie 1^@
so we find dv=-e.dr
therefore we know the potneital at the centre of the ring therefore work done =q change in potential
1@ all , i'm not able to get why the external agent will be doing any work, because as the metallic shell is neutral so it means that no electric field will be present in space( due to metallic shell) this means the agent does not have to do any work against any field, and potential at infinity = potential t centre of shell=0 , so W=0 , where am i wrong???
1@ Varun , but shell is neutral so potential at centre will be zero only, i agree that charges will be induced but still the potential at centre remains zero because net charge is equidisant ( from centre ) and zero in magnitude
1no ur mistaken the potneital at the centre will not be 0 use calculus to ifnd the correcr answer