# Find electric field strength expression for a specfic region

• jlmccart03
In summary: I think you're still getting tripped up by the volume calculation. I would expect the result to involve cubes in the numerator and a square in the denominator (corresponding to volume in the numerator as part of the calculation of the enclosed charge, and area in the denominator for the area of the...shell?).
jlmccart03

## Homework Statement

A thick, spherical shell of inner radius a and outer radius b carries a uniform volume charge density ρ.

Find an expression for the electric field strength in the region a<r<b.

## Homework Equations

Gauss's Law ∫E⋅dA
Area of a sphere

## The Attempt at a Solution

I know I am really close to the answer, but I do not know what I am doing wrong. I know that there are two radii, but how do I find the electric field strength with these. I think I am close, but may be way off too. Can someone guide me to the way of solving this.

What has been your approach so far? Can you show us an attempt?

gneill said:
What has been your approach so far? Can you show us an attempt?
Ok this is my attempt: (1/4πr^2)*(Qenclosed/ε0) = (1/4πr^2)*(πa^2/ε0) = (1/4r^2)(a^2/ε0). Correct?

How did you determine Qenclosed? I don't see the charge density ρ used, and a measure of volume should involve the cube of a radius (or two). Perhaps you should show in detail how you determined Qenclosed.

gneill said:
How did you determine Qenclosed? I don't see the charge density ρ used, and a measure of volume should involve the cube of a radius (or two). Perhaps you should show in detail how you determined Qenclosed.
Well, Qenclosed is simply πa^2ρ. I forgot to put in the ρ in the previous post. Its this because of the sphere. I just used a previous example where in class we used a cylinder instead and got it to be πr^2L for Qenclosed without the ρ. I may be confused though on this part.

The enclosed charge should go as the volume of the portion of the thick shell, so expect to see terms that are radius cubed, not squared. πa^2ρ having only a radius squared term indicates that it's an area element, not a volume. I think you need to think about how you might determine the volume of a portion of the shell.

gneill said:
The enclosed charge should go as the volume of the portion of the thick shell, so expect to see terms that are radius cubed, not squared. πa^2ρ having only a radius squared term indicates that it's an area element, not a volume. I think you need to think about how you might determine the volume of a portion of the shell.
Wait a minute. I thought I am dealing with area and not volume. Since I am dealing with Gauss's law which is ∫E⋅dA. I didn't think volume was a part of this question at all. Am I wrong?

jlmccart03 said:
Wait a minute. I thought I am dealing with area and not volume. Since I am dealing with Gauss's law which is ∫E⋅dA. I didn't think volume was a part of this question at all. Am I wrong?
You need the surface area of the Gaussian surface, and you also need the enclosed charge. The enclosed charge is spread over a volume, a portion of the thick shell. It's the subshell extending from radius ##a## to radius ##r##.

gneill said:
You need the surface area of the Gaussian surface, and you also need the enclosed charge. The enclosed charge is spread over a volume, a portion of the thick shell. It's the subshell extending from radius ##a## to radius ##r##.
Ok so in that case how do I apply the volume of a sphere being 4/3πr^3? Does the 4's cancel leaving (1/3r^3)(a^2/ε0ρ).

A shell is not a solid sphere, so its volume won't be that of a solid sphere.

First work out how to determine the volume of a spherical shell (the shell itself, not including the empty cavity that the shell encloses). The result will include variables a and r, with a being the inside surface radius of the shell and r some radius between a and b.

gneill said:
A shell is not a solid sphere, so its volume won't be that of a solid sphere.

First work out how to determine the volume of a spherical shell (the shell itself, not including the empty cavity that the shell encloses). The result will include variables a and r, with a being the inside surface radius of the shell and r some radius between a and b.
Ok so I tried my own way of getting the answer and came up with ((b2-a2)/4r3ε0))ρ.

It is incorrect. The units do not resolve to those of an electric field (N/C). They resolve to something like N/(C*m^2) instead.

I think you're still getting tripped up by the volume calculation. I would expect the result to involve cubes in the numerator and a square in the denominator (corresponding to volume in the numerator as part of the calculation of the enclosed charge, and area in the denominator for the area of the Gaussian surface).

Have you tried doing a web search to find the volume of a spherical shell so that you can see what you're trying to match?

gneill said:
It is incorrect. The units do not resolve to those of an electric field (N/C). They resolve to something like N/(C*m^2) instead.

I think you're still getting tripped up by the volume calculation. I would expect the result to involve cubes in the numerator and a square in the denominator (corresponding to volume in the numerator as part of the calculation of the enclosed charge, and area in the denominator for the area of the Gaussian surface).

Have you tried doing a web search to find the volume of a spherical shell so that you can see what you're trying to match?
So its V = (4/3)π(r3-(r-t)3). I need it in this form?

jlmccart03 said:
So its V = (4/3)π(r3-(r-t)3). I need it in this form?
Something like that, although I don't know what t is. It's not one of the variables defined in the problem.

I'll give you a hint: If you take the volume of a solid sphere and take away the volume of a slightly smaller sphere inside it (leaving an empty spherical cavity in its place), what sort of object do you end up with?

In this case the slightly smaller sphere represents the hollow cavity inside your shell (so radius a). The larger sphere represents the solid sphere out to the Gaussian surface that you've drawn within the original shell (radius r).

Last edited:
gneill said:
Something like that, although I don't know what t is. It's not one of the variables defined in the problem.

I'll give you a hint: If you take the volume of a solid sphere and take away the volume of a slightly smaller sphere inside it (leaving an empty spherical cavity in its place), what sort of object do you end up with?

In this case the slightly smaller sphere represents the hollow cavity inside your shell (so radius a). The larger sphere represents the solid sphere out to the Gaussian surface that you've drawn within the original shell (radius r).
So its something like (4/3)π(b3-(b-x)3)-(4/3)π(a3-(a-x3)? Its similar to a washer problem in Calculus correct?

It is similar, yes. But you don't need to introduce new variables. You already have the two radii, a and r.

gneill said:
It is similar, yes. But you don't need to introduce new variables. You already have the two radii, a and r.
But b is the outer radius and a is the inner radius, I actually have no clue where r is placed.

From the problem statement:

jlmccart03 said:
Find an expression for the electric field strength in the region a<r<b.

gneill said:
From the problem statement:
Sorry I didn't write it down, but the problem states to "Express your answer in terms of r, a, b, ρ, ϵ0." It was in the picture.

Also it says "A thick, spherical shell of inner radius a and outer radius b carries a uniform volume charge density ρ."

jlmccart03 said:
Sorry I didn't write it down, but the problem states to "Express your answer in terms of r, a, b, ρ, ϵ0." It was in the picture.

Also it says "A thick, spherical shell of inner radius a and outer radius b carries a uniform volume charge density ρ."
Yes, and r is in between a and b. a and r are the only two radii that are important (so long as r never exceeds b).

gneill said:
Yes, and r is in between a and b. a and r are the only two radii that are important (so long as r never exceeds b).
OHHHHH ok, so I just use a the inner radius and r the radius for the thick piece to find the expression. So once I subtract the two volumes I get the volume of the inner piece. From there do I do the basic Qenclosed/ε0 or is there another additional step?

No that's it. Just apply Gauss' Law to the enclosed charge.

gneill said:
No that's it. Just apply Gauss' Law to the enclosed charge.
So I should get (((r-a)x-r^2+b^2))x/ε0r^2)ρ. Correct?

No. there will be cubes in the numerator; You're dealing with volumes. And there will be no variable x. There will be variables a, r, ρ, and ε0 in the answer.

How are you calculating the volume of the shell of interest? Can you show your derivation?

gneill said:
No. there will be cubes in the numerator; You're dealing with volumes. And there will be no variable x. There will be variables a, r, ρ, and ε0 in the answer.

How are you calculating the volume of the shell of interest? Can you show your derivation?
The volume should actually be (4/3)π(b3-(b-r)3). Then I took that and applied (1/4πε0) to get r(r2-3br+3b2)/(3ε0)

jlmccart03 said:
The volume should actually be (4/3)π(b3-(b-r)3). Then I took that and applied (1/4πε0) to get r(r2-3br+3b2)/(3ε0)
I think you're choosing the wrong volume for the charge that's enclosed within the Gaussian surface. The variable b should not appear anywhere in the answer.

The shell you're interest in lies between radii a and r. It's labelled Qenc in the figure above. It is within the Gaussian surface, hence "enclosed".

gneill said:
I think you're choosing the wrong volume for the charge that's enclosed within the Gaussian surface. The variable b should not appear anywhere in the answer.

View attachment 112782

The shell you're interest in lies between radii a and r. It's labelled Qenc in the figure above. It is within the Gaussian surface, hence "enclosed".
Oh god I am so lost. Ok if I just switch b with a and flip a and r I should have the correct formula right?

jlmccart03 said:
Oh god I am so lost. Ok if I just switch b with a and flip a and r I should have the correct formula right?
No, because you've still not calculated a volume.

What is the volume of a sphere of radius r?

gneill said:
No, because you've still not calculated a volume.

What is the volume of a sphere of radius r?
4/3πr3

Right. And similarly, the volume of a sphere of radius a is (4/3)πr3.

The sphere of radius a happens to be the volume of the empty cavity within the shell. So the volume of the shell from a out to r is equal to the volume of the solid sphere of radius r minus the volume of the sphere of radius a (you're removing the volume of the cavity from that of a solid sphere).

gneill said:
Right. And similarly, the volume of a sphere of radius a is (4/3)πr3.

The sphere of radius a happens to be the volume of the empty cavity within the shell. So the volume of the shell from a out to r is equal to the volume of the solid sphere of radius r minus the volume of the sphere of radius a (you're removing the volume of the cavity from that of a solid sphere).
So ((4/3)π(r3-a3))/(4πε0))ρ.

jlmccart03 said:
So ((4/3)π(r3-a3))/(4πε0))ρ.
Close. You're missing the r2 in the denominator that belongs to the Gaussian surface area. Then you can cancel the 4π's that are common to the numerator and denominator.

gneill said:
Close. You're missing the r2 in the denominator that belongs to the Gaussian surface area. Then you can cancel the 4π's that are common to the numerator and denominator.
Ok so that would mean it becomes (r3-a3)/(3ε0r2))ρ?

Yes, that looks good.

gneill said:
Yes, that looks good.
Phew! Thanks for the help even though I got really lost, but I really appreciate you working through the roadblocks I was having! Thank you!

<h2>1) What is the formula for calculating electric field strength in a specific region?</h2><p>The formula for calculating electric field strength in a specific region is E = kq/r^2, where E is the electric field strength, k is the Coulomb's constant, q is the charge of the source, and r is the distance from the source.</p><h2>2) How does the shape of the region affect the electric field strength?</h2><p>The shape of the region can affect the electric field strength by changing the distribution of charges and altering the distance between the source and the point where the field is being measured. This can impact the value of r in the formula and ultimately change the magnitude of the electric field strength.</p><h2>3) Can the electric field strength in a specific region be negative?</h2><p>Yes, the electric field strength in a specific region can be negative. This indicates that the direction of the electric field is opposite to the direction of the force experienced by a positive test charge placed in that region. It is important to note that the magnitude of the electric field strength is always positive.</p><h2>4) How does the presence of multiple sources affect the electric field strength in a specific region?</h2><p>The presence of multiple sources can affect the electric field strength in a specific region by adding or subtracting their individual contributions. If the sources have the same charge, their electric fields will add up and result in a stronger field. If the sources have opposite charges, their electric fields will subtract and result in a weaker field.</p><h2>5) What factors can cause the electric field strength in a specific region to change?</h2><p>The electric field strength in a specific region can change due to changes in the charge of the source, changes in the distance between the source and the point where the field is being measured, and changes in the medium between the source and the point. Additionally, the presence of other charges or sources in the vicinity can also affect the electric field strength in a specific region.</p>

## 1) What is the formula for calculating electric field strength in a specific region?

The formula for calculating electric field strength in a specific region is E = kq/r^2, where E is the electric field strength, k is the Coulomb's constant, q is the charge of the source, and r is the distance from the source.

## 2) How does the shape of the region affect the electric field strength?

The shape of the region can affect the electric field strength by changing the distribution of charges and altering the distance between the source and the point where the field is being measured. This can impact the value of r in the formula and ultimately change the magnitude of the electric field strength.

## 3) Can the electric field strength in a specific region be negative?

Yes, the electric field strength in a specific region can be negative. This indicates that the direction of the electric field is opposite to the direction of the force experienced by a positive test charge placed in that region. It is important to note that the magnitude of the electric field strength is always positive.

## 4) How does the presence of multiple sources affect the electric field strength in a specific region?

The presence of multiple sources can affect the electric field strength in a specific region by adding or subtracting their individual contributions. If the sources have the same charge, their electric fields will add up and result in a stronger field. If the sources have opposite charges, their electric fields will subtract and result in a weaker field.

## 5) What factors can cause the electric field strength in a specific region to change?

The electric field strength in a specific region can change due to changes in the charge of the source, changes in the distance between the source and the point where the field is being measured, and changes in the medium between the source and the point. Additionally, the presence of other charges or sources in the vicinity can also affect the electric field strength in a specific region.

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