Sphere enclosed in shell is expanded, find work done

gnrlies00
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Homework Statement


Consider a non-conducting sphere of radius 'R' and charge 'Q' is enclosed by spherical shell of radius '5R' and charge '4Q'. If inner sphere is expanded to radius '3R'.Then amount of work done by the field in this process is

Homework Equations


W=ε0/2∫E2

The Attempt at a Solution


first find the energy of the non-conducting sphere of radius R and charge Q
W=ε0/2k2Q2(∫R 1/r4(r24π)dr + ∫0 R (r/R3)2 (4πr2)dr)
⇒ W=3kQ2 / 5R ---------------------------(1)

Similarly, find W for spherical shell
⇒ W=8kQ2/ 5R -------------------------(2)

Subtracting (2) from (1)
W= kQ2/5R = kQ2/R ----------------(3)

For the remaining part I'm confused...
Should I find the work done for the new expanded sphere(i.e. Radius = 3R) and subtract it from equation(3)..
If yes, then what limits should I take for the integral ∞ to 3R (for outside E) and 3R to 0 (for inside E)?

Am I going about the problem correctly or missing something?

Please help.
 
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gnrlies00 said:
Subtracting (2) from (1)
W= kQ2/5R = kQ2/R ----------------(3)

For the remaining part I'm confused...

Your confusion starts already before. You cannot compute the energies of the charge configurations separately as it is not linear in the field. Doing so neglects the potential between the configurations.
 
okay, does that mean I should find the potential difference between the sphere of radius R and spherical shell, and find the Energy.
then the potential of expanded sphere of radius 3R and spherical shell, find the energy.
and Difference in energy is the work done?
 
You can do it the way you indicated from the start, you just need to take the fields from both objects into account together.
 
potential on sphere of radius R
V=-kQ∫R1/r2dr = kQ/R ----------------(1)
potential on spherical shell is
V= 4kQ/5R -----------------(2)

(2)-(1)
V=kQ/5R--------------(3)

E=-∇V=kQ/5R2

⇒W=εo/2 ∫R5RE2
⇒W=εo/2 (kQ/5)2R5R 1/R2 4π dr = 2kQ2/5R ----------------(4)

Potential on sphere of radius 3R
V= kQ/3R -----------------(5)

(4)-(2)
V=7kQ/15R

E=-∇V=7kQ/15R2

W= εo/2 (7kQ/15)23R5R 1/R2 4π dr ----------------(6)

(6)-(4) is the answer

is this correct, I don't feel I'm in the right direction.




 
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