Gauss's Law for a sphere with a cavity, solving for E(r)

[ThrawnGaming]

Homework Statement

This is for my AP Physics E & M class.

Relevant Equations

closed integral(E dot dA) = qenclosed/eps0

I am not sure how to solve for E(r) for R1<r<R2.

 

[vela]

Try using Gauss's Law.

 

[Gordianus]

You should start by showing us your previous attempt. What have you found so far?

 

[PhDeezNutz]

Can you find the total enclosed charge for spherical surface of radius $r$ where $R_1 \lt r \lt R_2$.

Hint pretend there is no cavity (i.e. the cavity is the same density $\rho = 715 \frac{\mu C}{m^3}$ as the rest of the sphere and then subtract the charge of the filled in cavity).

Or if you know integration just spherically integrate from $R_1$ to $r$

 

[ThrawnGaming]

You should start by showing us your previous attempt. What have you found so far?

I tried integrating for 4/3 pi r^3 with r= 0.04 to r=0.12, and I found that the charge decayed exponentially from r=0.04 (1/r^2), but I know that is incorrect, because for volume charge densities, it should increase from r=0.04...

 

[vela]

We'd like to see the actual math as verbal descriptions are often inaccurate and misleading. From what you described, you went from something wrong to something even more wrong. I have no idea what "charge decayed exponentially from r=0.04 (1/r^2)" actually means.

If you haven't already, please read https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/ to avoid making common mistakes when asking for assistance here.

 

[ThrawnGaming]

 

[PhDeezNutz]

I tried integrating for 4/3 pi r^3 with r= 0.04 to r=0.12, and I found that the charge decayed exponentially from r=0.04 (1/r^2), but I know that is incorrect, because for volume charge densities, it should increase from r=0.04...

Is this AP physics B or C? Have you taken multi variable calculus? If not don’t worry about computing the integral explicitly, just use the idea of subtracting a sphere from a bigger sphere to find the volume (and therefore total charge) of a thick spherical shell.

 

[ThrawnGaming]

Is this AP physics B or C? Have you taken multi variable calculus? If not don’t worry about computing the integral explicitly, just use the idea of subtracting a sphere from a bigger sphere to find the volume (and therefore total charge) of a thick spherical shell.

This is AP Physics C, and I am currently taking AP Calculus AB

 

[ThrawnGaming]

This is AP Physics C, and I am currently taking AP Calculus AB

I want to find E as a function of r, and I think I have to integrate to do that.

 

[PhDeezNutz]

View attachment 294042

You’re integrating the expression for volume of a sphere you’re not integrating the expression that leads to the expression for volume of a sphere.

$dV$ is actually $r^2 \sin \theta dr d\theta d\phi$ in spherical coordinates. If you integrate that over $0 \lt r \lt R$, $0 \lt \theta \lt \pi$, and $0 \lt \phi \lt 2 \pi$ you will get $\frac{4}{3}\pi r^3$.

But you only want to integrate from $R_1$ to $r$ right?

Again if you haven’t taken Multivariable Calculus subtract the small sphere from the bigger sphere to “cut out a cavity”.

 

[PhDeezNutz]

I want to find E as a function of r, and I think I have to integrate to do that.

Well integration and my suggestion will have the same effect.

I’m trying to help you without breaking the forum rules and giving you a solution.

Say you wanted to find the total charge out to $8 cm$? You’d find the enclosed charge of a solid sphere that has a radius of $8 cm$ and subtract the enclosed charge of a sphere with a radius of $4 cm$.

Now replace $8 cm$ with “r” and isn’t the problem exactly the same?

Just qualify if r is less than 4 the field is zero.

 

[ThrawnGaming]

Well integration and my suggestion will have the same effect.

I’m trying to help you without breaking the forum rules and giving you a solution.

Say you wanted to find the total charge out to $8 cm$? You’d find the enclosed charge of a solid sphere that has a radius of $8 cm$ and subtract the enclosed charge of a sphere with a radius of $4 cm$.

Now replace $8 cm$ with “r” and isn’t the problem exactly the same?

Just qualify if r is less than 4 the field is zero.

Is this correct?

 

[PhDeezNutz]

I can’t make everything out but from the last line it doesn’t look to be.

We have constant density $\rho$ correct? Charge is volume times density…correct?

Now to find total charge we have to find the volume. If you’re dead set on integrating on the right hand side you would integrate the surface area (times $dr$) to find volume. You don’t integrate volume to find volume. Volume is the integral of surface area.

Compute $\rho \int_{R_1}^{r} 4 \pi r^2 \, dr$.

Set that equal to $E (4 \pi r^2 )$ and solve for $E$

 

[ThrawnGaming]

I can’t make everything out but from the last line it doesn’t look to be.

We have constant density $\rho$ correct? Charge is volume times density…correct?

Now to find total charge we have to find the volume. If you’re dead set on integrating on the right hand side you would integrate the surface area (times $dr$) to find volume. You don’t integrate volume to find volume. Volume is the integral of surface area.

Compute $\rho \int_{R_1}^{r} 4 \pi r^2 \, dr$.

Set that equal to $E (4 \pi r^2 )$ and solve for $E$

Thank you soooo much! I think I understand now. I have $\rho(4/3 \pi r^3 - 4/3 \pi R_1^3)$

 

[PhDeezNutz]

Thank you soooo much! I think I understand now. I have $\rho(4/3 \pi r^3 - 4/3 \pi R_1^3)$

You’re halfway there. Now finish the problem.

Find the field in all 3 regions.

In the cavity.

In the shell.

Outside the entire sphere.