# Sphere enclosed in shell is expanded, find work done

• gnrlies00
In summary, the problem involves finding the amount of work done by the electric field when a non-conducting sphere of radius R and charge Q is expanded to a radius of 3R while enclosed by a spherical shell of radius 5R and charge 4Q. The correct approach is to take into account the potential between the two charge configurations and not to compute their energies separately. The final answer can be obtained by subtracting the potential on the expanded sphere from the potential on the original sphere and using the resulting field to calculate the work done by the field.
gnrlies00

## 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

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?

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)

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

## 1. What does it mean for a sphere to be enclosed in a shell?

When we say that a sphere is enclosed in a shell, it means that the sphere is surrounded by another, larger object that completely encompasses it. Think of it like a ball inside a bigger ball.

## 2. How is the expansion of the sphere related to the work done?

The expansion of the sphere is directly related to the work done on it. As the sphere expands, work is done by external forces to move the shell outward and expand its volume.

## 3. What factors determine the amount of work done on the sphere?

The amount of work done on the sphere depends on the initial and final volumes of the sphere, the pressure applied to the shell, and the elasticity of the shell material.

## 4. How is the work done calculated for a sphere enclosed in a shell?

The work done on a sphere enclosed in a shell can be calculated using the formula W = P * (V2 - V1), where W is the work done, P is the pressure applied, and V1 and V2 are the initial and final volumes of the sphere, respectively.

## 5. Can the work done on a sphere enclosed in a shell be negative?

Yes, the work done on a sphere enclosed in a shell can be negative if the sphere is compressing instead of expanding. In this case, work is done on the sphere by external forces to decrease its volume.

Replies
2
Views
2K
Replies
29
Views
3K
Replies
4
Views
3K
Replies
1
Views
2K
Replies
1
Views
2K
Replies
7
Views
4K
Replies
49
Views
5K
Replies
4
Views
4K
Replies
2
Views
2K
Replies
6
Views
4K