Electric field in a spherical shell

In summary: This is done by choosing a value of A that makes the integral of the electric field between a and b vanish.
  • #1
259
11
Homework Statement
There is given a spherical shell with an inner radius of a and an outer radius of b, and there is a point charge +Q in the center. The volume charge density of the charge in the shell is A/r where A is some constant.

Find A if the electric field inside the shell is constant.
Relevant Equations
Gauss' law
Hi, been a while since I last asked here something.

I am restudying electrostatics right now, and I am facing difficulties in the following question:

My attempt:

I tried to use Gauss' law, what I got is the equation in the capture but that doesn't lead me anywhere as I am unable to find a value that will result in a constant field inside the shell. What am I doing wrong?

Of course that the equation is for a<r<b
 

Attachments

  • Capture.PNG
    Capture.PNG
    2 KB · Views: 173
Physics news on Phys.org
  • #2
Eitan Levy said:
what I got
And how, precisely, did you get that ? (you want to know what you did wrong, so we want to know what you did. But all you telll us is the outcome -- which is wrong - or not)

My clues:
  1. for QQ alone the field is ##kQ\over r^2##kQr2 and not 4π4π times as much
  2. you multiply ##A\over r## with a volume -- that's not Gauss !
 
  • #3
BvU said:
And how, precisely, did you get that ? (you want to know what you did wrong, so we want to know what you did. But all you telll us is the outcome -- which is wrong - or not)
Simply by looking at a sphere with a radius of r. The total charge inside the surface of this sphere should amount to Q, the point charge plus the charge in the shell until we reach a distance of r. Then I simply used the formula for Gauss' law.
 
  • #4
(I give up editing my post #2 - PF is whacking it all the time)

Eitan Levy said:
he charge in the shell until we reach a distance of r
How do you calculate that ?
 
  • #5
BvU said:
(I give up editing my post #2 - PF is whacking it all the time)

How do you calculate that ?

A massive error by me, thank you.
 
  • #6
BvU said:
(I give up editing my post #2 - PF is whacking it all the time)

How do you calculate that ?
Actually I am still unable to reach a solution.

I calculated that by calculating the integral of A/r*4πr^2dr from a to r, I got 2πA(r^2-a^2) as an answer.

EDIT: Never mind I got it, thanks.
 
  • #7
Eitan Levy said:
Homework Statement: There is given a spherical shell with an inner radius of a and an outer radius of b, and there is a point charge +Q in the center. The volume charge density of the charge in the shell is A/r where A is some constant.

Find A if the electric field inside the shell is constant.
You may be misinterpreting the question.
A spherically symmetric charge generates no field inside itself. They must mean the field is constant between the inner and outer radii of the shell.
This does not require integration. Consider a point at radius r between a and b. What is the field there due to:
- the point charge
- the spherical charge between a and r
- the spherical charge between r and b?
 
  • Like
Likes collinsmark
  • #8
haruspex said:
the spherical charge between a and r
requires integration ...
 
  • #9
BvU said:
requires integration ...
:smile: Yes, that's true. (integration is required to find the charge inside the shell; or even the charge inside the shell from radius [itex] a [/itex] to [itex] r [/itex]).

But I must be misinterpreting the problem. I can't seem to find a solution where both [itex] E [/itex] and [itex] A [/itex] are constants (for any [itex] r [/itex] that lies inside the spherical shell). I attribute that to my misunderstanding of something, somewhere or another. 🤔
 
  • #10
BvU said:
requires integration ...
It can be avoided. Consider the rate of change of the field as radius increases at radius a.
For the field due to the point charge that rate is ##-\frac{2Q}{a^3}##. For the spherical shell it is ##4\pi\frac Aaa^2\frac 1{a^2}=\frac{4\pi A}a##.
The sum of the two must be zero.
Arguably, this approach assumes there is an answer.
 
Last edited:
  • #11
collinsmark said:
But I must be misinterpreting the problem. I can't seem to find a solution where both [itex] E [/itex] and [itex] A [/itex] are constants (for any [itex] r [/itex] that lies inside the spherical shell). I attribute that to my misunderstanding of something, somewhere or another. 🤔
Nevermind. I see my misunderstanding now. :smile: For a specific value of A (depending on other paramters such as Q and a), both A and E can be made to be constants for the region between a and b.
 
Last edited:

1. What is an electric field in a spherical shell?

The electric field in a spherical shell is the force exerted per unit charge at any point outside the shell due to the distribution of charges on the surface of the shell.

2. How is the electric field in a spherical shell calculated?

The electric field in a spherical shell is calculated by using the formula E = kQ/r², where E is the electric field, k is the Coulomb's constant, Q is the total charge on the shell, and r is the distance from the center of the shell to the point where the electric field is being measured.

3. Is the electric field in a spherical shell uniform?

Yes, the electric field in a spherical shell is uniform at all points outside the shell. This means that the magnitude and direction of the electric field are the same at all points.

4. What happens to the electric field inside a spherical shell?

Inside a spherical shell, the electric field is zero. This is because the charges on the surface of the shell cancel each other out, resulting in a net electric field of zero.

5. How does the electric field in a spherical shell change with distance?

The electric field in a spherical shell follows an inverse-square relationship with distance. This means that as the distance from the shell increases, the electric field decreases proportionally.

Suggested for: Electric field in a spherical shell

Replies
9
Views
543
Replies
9
Views
1K
Replies
2
Views
784
Replies
20
Views
394
Replies
28
Views
407
Back
Top