Charge density and charge on a conductor

[doggydan42]

Homework Statement

A point charge $q = −5.0\times 10^{−12} C$ is placed at the center of a spherical conducting shell of inner radius 3.5 cm and outer radius 4.0 cm. The electric field just above the surface of the conductor is directed radially outward and has magnitude 8.0 N/C. (a) What is the charge density on the inner surface of the shell? (b) What is the charge density on the outer surface of the shell? (c) What is the net charge on the conductor?

Homework Equations

$$\vec E = \frac{\sigma}{\epsilon_0}
\\ \sigma = \vec E \epsilon_0 = \frac{q}{4\pi r^2}$$

The Attempt at a Solution

I originally plugged in the charge and the radii into the equation to find the charge density of the inner and outer surfaces. In that case $r=.035 m$ for the inner surface, and $r=.035+.04 m = .075 m$ for the outer surface. This resulted in $\sigma_{inner}=-3.3\times 10^{-10} \frac{C}{m^2}$ and $\sigma_{outer}=-7.1\times 10^{-11} \frac{C}{m^2}$. When just using the electric field for the outer charge density, The result is $\sigma_{outer}=-7.1\times 10^{-11} \frac{C}{m^2}$

Since the sphere is a conductor and E is the charge on the surface of a conductor, which is the same when using the radius, would the charge of the inner surface equal 0? Also, would the net charge on the conductor just be $q=4\pi r_{outer}^2\sigma$?

Thank you in advance.

 

[NFuller]

and $r=.035+.04 m = .075 m$ for the outer surface

The problem says the radius of the outer surface is 0.04m.

E is the charge on the surface of a conductor

$E$ is not the charge on the surface, they are two different things. Use conservation of charge to answer the last part.

 

[doggydan42]

$E$ is not the charge on the surface, they are two different things.

Sorry, I meant that E was the electric field on the surface, not the charge.

The problem says the radius of the outer surface is 0.04m.

That is what I originally thought, but when you use the electric field, E, instead of the charge and the radius, you get a different answer. On the other hand, when you take the summation for the radii, you get the same answer for the electric field on the outer surface.

Use conservation of charge to answer the last part.

So the charge on the surface for the conductor would remain $q=-5\times 10^{-12} C$? If so, what about the charge on the inner radius, should that not effect the answer?

 

[NFuller]

That is what I originally thought, but when you use the electric field, E, instead of the charge and the radius, you get a different answer. On the other hand, when you take the summation for the radii, you get the same answer for the electric field on the outer surface.

I'm not sure what you mean here. We are not allowed to change the radius.

So the charge on the surface for the conductor would remain q=−5×10−12Cq=−5×10−12Cq=-5\times 10^{-12} C? If so, what about the charge on the inner radius, should that not effect the answer?

How much charge did the conductor have initially before the charge was inserted in the center of the sphere?

 

[doggydan42]

I'm not sure what you mean here. We are not allowed to change the radius.

To clarify, I meant that if they gave the outer radius as a "layer" above the inner surface. So the distance from the inner surface to the outer surface was what they called the outer radius. I did that just as a test to see how it affected my results. What I found was that if I use that, so the radius from the center to the outer surface is $.035+.04 m$ then the result for the charge density is the same as using the electric field. If you use just .04 m as the radius of the outer surface, the result is different. I wasn't sure why there would be such a large difference depending if I solve for sigma in terms of the radius and charge, versus using the electric field just above the surface.

How much charge did the conductor have initially before the charge was inserted in the center of the sphere?

The charge would be 0 without the charge, right? If not, then would the method to solve for it be using the electric field to find the charge on the surface, and then subtracting the charge that can be found when using the radius of .04 m.
So,
$$q_2=A\epsilon_0 E
\\ q_1=\sigma A
\\q_{net charge}=q_2-q_1$$

 

[NFuller]

To clarify, I meant that if they gave the outer radius as a "layer" above the inner surface. So the distance from the inner surface to the outer surface was what they called the outer radius. I did that just as a test to see how it affected my results. What I found was that if I use that, so the radius from the center to the outer surface is .035+.04m.035+.04m.035+.04 m then the result for the charge density is the same as using the electric field. If you use just .04 m as the radius of the outer surface, the result is different. I wasn't sure why there would be such a large difference depending if I solve for sigma in terms of the radius and charge, versus using the electric field just above the surface.

I'm not sure how you can test this since the electric field outside the conductor will be the same regardless of the radius of the outer surface. The wording of the problem seems to explicitly state the outer radius as 0.4m so this is what we need to use.

The charge would be 0 without the charge, right?

Right, so $q_{1}+q_{2}=0$ where $q_{1}$ and $q_{2}$ are the charges on the inner and outer surfaces. The other important piece of information is that the electric field inside the conducting material is zero. Using this what is $\sigma_{1}$ and $\sigma_{2}$?

 

[doggydan42]

Right, so q1+q2=0q_{1}+q_{2}=0 where q1q_{1} and q2q_{2} are the charges on the inner and outer surfaces. The other important piece of information is that the electric field inside the conducting material is zero. Using this what is σ1\sigma_{1} and σ2\sigma_{2}?

If the inside of the conducting material has no net charge, and the inner charge density is $\sigma_1$, would the $\sigma_1$ not be 0?
With a radius of .04 m, the outer charge density becomes:
$$\sigma_2 = \frac{q}{4\pi r^2} = \frac{-5.0 \times 10^{-12} C}{4\pi (.04 m)^2} =-2.5\times 10^{-10} \frac{C}{m^2}$$

For the inner charge density, it would become:
$$\sigma_1 = \frac{q}{4\pi r^2} = \frac{-5.0 \times 10^{-12} C}{4\pi (.035 m)^2} = -3.2\times 10^{-10} \frac{C}{m^2}$$
Should $\sigma_1$ be positive so that $q_1+q_2=0$?

So the net charge on the conductor is 0?
Does this mean that the electric field was not an important part in the calculation, or am I missing a part?

Thank you

 

[NFuller]

I'm sorry, I just realized I made a mistake here. The initial charge on the sphere is not zero, that's why they gave you the electric field. This can be shown with Gauss's law at the surface where they gave the electric field as 8.0N/C directed outwards.
$$EA=\frac{Q_{total}}{\epsilon_{0}}$$
$$E4\pi r_{2}^{2}\epsilon_{0}=Q_{total}=q+q_{shell}$$
$$q_{shell}=4\pi r_{2}^{2}\epsilon_{0}E-q$$

Should σ1σ1\sigma_1 be positive so that q1+q2=0q1+q2=0q_1+q_2=0?

The interior surface charge can be found by applying Gauss's law inside the conducting shell where the electric field is zero. The exterior surface charge can be found using conservation of charge $q_{1}+q_{2}=q_{shell}$.

 

[doggydan42]

The interior surface charge can be found by applying Gauss's law inside the conducting shell where the electric field is zero. The exterior surface charge can be found using conservation of charge q1+q2=qshellq_{1}+q_{2}=q_{shell}.

Is $q_{shell}$ the net charge on the conductor?

 

[NFuller]

Is $q_{shell}$ the net charge on the conductor?

Yes

 

[doggydan42]

This can be shown with Gauss's law at the surface where they gave the electric field as 8.0N/C directed outwards.

Okay, so that would mean:

$$q_{net}=A_2\epsilon_0 E - q = 4\pi (.04m)^2(8.85\times 10^{-12} \frac{C^2}{Nm^2}(8.0 N/C)-5.0\times 10^{-12}=-3.6\times 10^{-12}$$

Is that right? Also, did I get the charge density right for the inner and outer surfaces?

Thank you.

 

[NFuller]

Is that right? Also, did I get the charge density right for the inner and outer surfaces?

This looks right. The charge density you calculated earlier is different now since we are not assuming the shell was uncharged.

 

[doggydan42]

The charge density you calculated earlier is different now since we are not assuming the shell was uncharged.

So would this be right then?
$$q_2=4\pi r_2^2 E\epsilon_0 \Rightarrow \sigma_2 = \frac{q_2}{A_2} = E\epsilon_0$$

Then it would follow that from $q_1+q_2 = q_{net}$ That we can solve for $q_1$, and so $\sigma_1 = q_1/A_1$
If I did this correctly, then $\sigma_2=7.08\times 10^{-11} \frac{C}{m^2}$, and from $q_1=5\times 10^{-12} C$, $\sigma_1=3.2\times 10^{-10} \frac{C}{m^2}$

And $q_{net}=6.4\times 10^{-12} C$

Thank you.

 

[NFuller]

I think your logic is right here but I just noticed you made a sign mistake in post #11 where you calculated $q_{net}$.
$q=-5.0\times 10^{-12}C$ so $-q=+5.0\times 10^{-12}C$.

 

[doggydan42]

I think your logic is right here but I just noticed you made a sign mistake in post #11 where you calculated qnetq_{net}.
q=−5.0×10−12Cq=-5.0\times 10^{-12}C so −q=+5.0×10−12C-q=+5.0\times 10^{-12}C.

Yes, thank you. I noticed that and corrected it.

So I understand the math and some of the equations, but why does $q_2=q_{tot}$, and why does $q_2 = q+q_{net}$? Though I do understand why $q_1+q_2=q_{net}$

Thank you.

 

[NFuller]

The electric field inside the conductor is zero. The only way this happens is if the inside charges $q$ and $q_{1}$ cancel out. The outside electric field must satisfy Gauss's law but now the only charge contributing to this field is $q_{2}$. Thus $q_{2}$ is the sum of both $q$ and $q_{net}$.