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1. Homework Statement
A solid conducting sphere of radius 'a' having a charge q is surrounded by a concentric conducting sphereical shell of inner radius 2a and outer radius 3a (please refer diagram)....
Find the amount of heat produced when switch is closed ....
2. Homework Equations
3. The Attempt at a Solution
I proceeded by finding out the Potential energy(electrostatic) of the system initially and finally and their difference is equal to heat lost....Doing all the calculations i get kq^{2}/2a as theanswer but answer is kq^{2}/4a......
So can anyone try and tell me if my answer is correct or not.....
I proceeded as below......
part 1.(before switch is closed)
charge distribution is q in the outer surface of smaller sphere, q on the inner surface of shell and q on the outer surface of shell.....PE of system at this stage is 5kq^{2}/6a.
Part 2(after switch is closed)
charge distribution is 0 on the outer surface of smaller sphere and 0 also on the inner surface of shell and q on the outer surface of shell.....
PE of system at this stage is kq^{2}/3a
Difference between Pe = heat so answer is kq^{2}/2a....
A solid conducting sphere of radius 'a' having a charge q is surrounded by a concentric conducting sphereical shell of inner radius 2a and outer radius 3a (please refer diagram)....
Find the amount of heat produced when switch is closed ....
2. Homework Equations
3. The Attempt at a Solution
I proceeded by finding out the Potential energy(electrostatic) of the system initially and finally and their difference is equal to heat lost....Doing all the calculations i get kq^{2}/2a as theanswer but answer is kq^{2}/4a......
So can anyone try and tell me if my answer is correct or not.....
I proceeded as below......
part 1.(before switch is closed)
charge distribution is q in the outer surface of smaller sphere, q on the inner surface of shell and q on the outer surface of shell.....PE of system at this stage is 5kq^{2}/6a.
Part 2(after switch is closed)
charge distribution is 0 on the outer surface of smaller sphere and 0 also on the inner surface of shell and q on the outer surface of shell.....
PE of system at this stage is kq^{2}/3a
Difference between Pe = heat so answer is kq^{2}/2a....
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