Electrostatics: "Compressing" Charged Spherical Shell

  • #1
479
12

Homework Statement


  1. How much work is required to squeeze a uniformly charged spherical shell from a radius of ##r## to a radius of ##r−dr##, if
    (a) the total charge q is a constant,
    (b) the sphere is kept at a constant potential, e.g. grounded.
    (c) Are the answers the same or different? Explain.

Homework Equations




The Attempt at a Solution



I'd like some help in checking that I have applied the right concept to the problem in my solution, for part a) first at least.

There is electrostatic pressure acting on the charges on the sphere's surface. To squeeze these charges in , one has to overcome such this force.

The electrostatic pressure is
$$F/A = \frac{\epsilon _0}{2} E^2$$
The electric-field magnitude on the sphere is
$$E = k\frac{q}{r^2}$$
The force causing such a pressure is
$$F = 2\pi r^2 \epsilon _0 E^2$$
$$F = \frac{1}{8\pi \epsilon _0}\frac{q^2}{r^2}$$
And the infinitesimal amount of work needed to squeeze these charges by ##dr## is
$$W_{on} = -W_{by} = -\frac{1}{8\pi \epsilon _0}\frac{q^2}{r^2} dr$$

Does any of this make sense? Thanks in advance!
 

Answers and Replies

  • #2
963
214
The electric-field magnitude on the sphere is
E=kqr2​

pl. check this expression ...on the surface of the shell it should be 2. PI. surface charge density i.e equivalent to (1/2).q/r^2...
 
  • #3
479
12
pl. check this expression ...on the surface of the shell it should be 2. PI. surface charge density i.e equivalent to (1/2).q/r^2...


I'm not sure what you mean. Why is E-field due to a spherical conducting shell ##\frac{q}{2r^2}##? Where does the ##\epsilon _0## go as well?

Isn't the surface charge density ##\sigma = \frac{q}{4\pi r^2}##, which means, if we write it in terms of ##\sigma##, ##E = \sigma / \epsilon _0##?

Thanks for assisting.
 
  • #4
963
214
I'm not sure what you mean. Why is E-field due to a spherical conducting shell q2r2q2r2\frac{q}{2r^2}?

i meant to say that the electric field "on the surface "should be " half of q / epsilon zero " and in Gaussian units we write it as 2.pi. charge density. however outside the surface it can be your expression i.e. k.q/ r^2....
 
  • #5
479
12
i meant to say that the electric field "on the surface "should be " half of q / epsilon zero " and in Gaussian units we write it as 2.pi. charge density. however outside the surface it can be your expression i.e. k.q/ r^2....

I think I get what you mean. In SI units, the field on the surface of a conductor is ##\frac{\sigma}{2\epsilon _0}## which I should have written instead of the usual ##\frac{kq}{r^2}## just outside the surface.

However, the equation I use later for electrostatic pressure ##P = \frac{\epsilon _0}{2} E^2## is written in terms of the field just outside the surface is it not?

If so, does the rest of my steps flow correctly?

Thank you for your assistance!
 
  • #6
963
214
If so, does the rest of my steps flow correctly?

The amount of work done in sqeezing is equivalent to (Pressure.x surface area ). dr = (1/2 ). (epsilonzero)E^2 ). 4. pi r^2 dr so i think it is correct. sorry for the intrusion.
 
  • #7
479
12
The amount of work done in sqeezing is equivalent to (Pressure.x surface area ). dr = (1/2 ). (epsilonzero)E^2 ). 4. pi r^2 dr so i think it is correct. sorry for the intrusion.

Thank you for the assistance!

For part b) it was mentioned that the sphere was grounded. Because ##\vec{E} = -\nabla V##, would I be right to say that constant ##V## means constant ##E##? Which means that I follow through the same steps as part a), except that I always take E to be the field strength just outside the sphere of initial radius ##r##?
 
  • #8
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,175
6,794
For part b) it was mentioned that the sphere was grounded.
No, it says "e.g. grounded", but more generally it is simply a constant potential.
would I be right to say that constant V means constant E?
Can you justify that? What is the relationship between E, V and r?

For both parts of the problem, I would feel safer starting with (or deriving) the expression for energy embodied in a sphere of given radius and charge, then seeing how that changes as the radius changes. For part b, you need to express the charge as a function of r and potential first.

Hint: be careful with signs.
 
  • #9
479
12
Thank you for your response.

Can you justify that? What is the relationship between E, V and r?

Wait, in a region of constant potential, there isn't any ##E## field, was that my error?

Also may I try to re-state my understanding of the second problem so that you can correct me?

Initial potential on the surface of the shell is given by $$V_{0} = k \frac{q_0}{r_0}$$
If potential is kept constant, ##q## varies according to $$q = \frac{V_0}{k} r$$
Field just outside the surface of the sphere is $$E_{out} = k \frac{q}{r^2}$$
which under constant potential varies with ##r## as follows $$E_{out} = V_0/r $$
Electrostatic Pressure on the charges is $$P =\frac{1}{2} \epsilon _0 E_{out} ^2$$
Electrostatic Force on the charges is $$F = 2\pi \epsilon _0 V_0 ^2 $$
And work done to overcome such a force through ##dr## is $$W = -\frac{1}{8\pi \epsilon _0} \frac{q_0 ^2}{r_0 ^2}dr$$

Is this right?

For both parts of the problem, I would feel safer starting with (or deriving) the expression for energy embodied in a sphere of given radius and charge, then seeing how that changes as the radius changes. For part b, you need to express the charge as a function of r and potential first.

Do you mean that I could also consider the energy needed to form spheres of radius ##r## and ##r - dr##, and take the difference between them as the work done?
 
  • #10
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,175
6,794
Do you mean that I could also consider the energy needed to form spheres of radius r and r−dr, and take the difference between them as the work done?
Yes, but it looks like you got the same result as I did already.
What do you notice about how this answer compares with that for part a)?
 
  • #11
479
12
Yes, but it looks like you got the same result as I did already.
What do you notice about how this answer compares with that for part a)?

The force applied in the first part is dependent on varying radius ##r##. The force applied in the second part is dependent only on initial radius ##r_0## and is therefore a constant.

Is this right?

Thank you!
 
  • #12
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,175
6,794
The force applied in the first part is dependent on varying radius ##r##. The force applied in the second part is dependent only on initial radius ##r_0## and is therefore a constant.

Is this right?

Thank you!
It is only a small change, dr, so that will not make much difference. There is a much more significant difference. What was my hint in post #8?
 
  • #13
479
12
It is only a small change, dr, so that will not make much difference. There is a much more significant difference. What was my hint in post #8?

You said to watch the signs, but I can't seem to connect the dots. What am I not seeing?
 
  • #14
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,175
6,794
You said to watch the signs, but I can't seem to connect the dots. What am I not seeing?
In post #1 you got a negative value for the work done. Does that make sense for part a)?
 
  • #15
479
12
In post #1 you got a negative value for the work done. Does that make sense for part a)?

Since motion is in the direction of the applied force, I guess not. But doesn't this mean part b) should change signs as well?

Thank you!
 
  • #16
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,175
6,794
doesn't this mean part b) should change signs as well?
Does it? Can you find a step where the sign went wrong in part a)? Can you find a step where you have a sign error in part b)?
 
  • #17
479
12
Does it? Can you find a step where the sign went wrong in part a)? Can you find a step where you have a sign error in part b)?

Apologies, but I am still unable to follow. I think it's best I spell out my train of thought.

There is a force pointing radially outwards of the spherical shell. If I want to squeeze this shell inwards, I have to overcome this force over a change in radius ##dr##, which inherently has a negative sign. Work done by the electrostatic force is ##Fdr##, which is negative. Work done by me is there for ##-Fdr##

Since I need to overcome similar forces in both parts of the problem, they "should" have the same sign. What's missing from this argument?

Thank you for your patience!
 
  • #18
TSny
Homework Helper
Gold Member
13,096
3,416
I don't know if this is part of the sign confusion. The problem statement says
How much work is required to squeeze a uniformly charged spherical shell from a radius of ##r## to a radius of ##r−dr##,
Note that the subtraction ##r - dr## means that the problem statement is taking ##dr## to be positive even though the radius decreases. :oldgrumpy:

It might be less confusing to use absolute values, so that the final radius is ##r - |dr|##. And, to avoid ambiguity in the signs of the answers for the work, express the work in terms of ##|dr|## instead of ##dr##.
 
  • #19
479
12
I don't know if this is part of the sign confusion. The problem statement says
Note that the subtraction ##r - dr## means that the problem statement is taking ##dr## to be positive even though the radius decreases. :oldgrumpy:

It might be less confusing to use absolute values, so that the final radius is ##r - |dr|##. And, to avoid ambiguity in the answers for the work, express the work in terms of ##|dr|## instead of ##dr##.

Oops. Actually lost track of that. In this case, would both cases of ##W## be in the form ##Fdr##?

Thank you for pointing that out.
 
  • #20
TSny
Homework Helper
Gold Member
13,096
3,416
Oops. Actually lost track of that. In this case, would both cases of ##W## be in the form ##Fdr##?
Yes, if ##F## is the magnitude of the force (so, a positive quantity) and if ##dr## is the magnitude of the displacement (so, also positive).
 
  • #21
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,175
6,794
Yes, if ##F## is the magnitude of the force (so, a positive quantity) and if ##dr## is the magnitude of the displacement (so, also positive).
Yes, that would explain the wrong sign in part a), but I am uneasy with the method, and in even more so when applying it to part b). It doesn't seem to account for the fact that the charges will flow off the sphere into the constant potential reservoir.
@WWCY, please try my method comparing the energies in the spheres before and after. The charges that flow off the spheres should do so with no change in energy.
 
  • #22
TSny
Homework Helper
Gold Member
13,096
3,416
Yes, that would explain the wrong sign in part a), but I am uneasy with the method, and in even more so when applying it to part b). It doesn't seem to account for the fact that the charges will flow off the sphere into the constant potential reservoir.

The work done by the "compressor" should be ##|F dr|## for both parts (a) and (b). This way, there is no need to consider potential energy.

If you want to relate the work done by the compressor to the change in potential energy of the system, then in part (b) you will need to take into account the change in potential energy of the reservoir.
 
  • #23
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,175
6,794
The work done by the "compressor" should be |F dr|
Why the absolute value? In principle, work done can be negative.
If you want to relate the work done by the compressor to the change in potential energy of the system, then in part (b) you will need to take into account the change in potential energy of the reservoir.
Ah, true, and I believe that leads to the same answer as in a).
Interesting; intuitively I would have said less work was involved in b), but it appears that it is the same to the first order of approximation, less to a second order.
 
  • #24
TSny
Homework Helper
Gold Member
13,096
3,416
Why the absolute value? In principle, work done can be negative.
Yes, but in this problem it’s clear that the work is positive when r decreases.The absolute value just makes sure you get a positive work whether you take dr as positive or negative. See the discussion in post #18 regarding the sign of dr.
 
  • #25
479
12
Yes, that would explain the wrong sign in part a), but I am uneasy with the method, and in even more so when applying it to part b). It doesn't seem to account for the fact that the charges will flow off the sphere into the constant potential reservoir.

May I know what you mean when you say "uneasy"? Why would explicitly calculating work done in such a manner be risky?

please try my method comparing the energies in the spheres before and after

Do I do this by using say, ##W = \frac{1}{2} \int_{s} \sigma V \ da## to calculate potential energy of a sphere with radius ##r##, then ##r-dr##, then taking the their difference (in both parts)?

Thank you!
 

Related Threads on Electrostatics: "Compressing" Charged Spherical Shell

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
4
Views
2K
Replies
1
Views
2K
  • Last Post
Replies
3
Views
5K
  • Last Post
Replies
1
Views
901
Replies
9
Views
3K
  • Last Post
Replies
16
Views
303
Replies
5
Views
3K
  • Last Post
Replies
4
Views
780
Top