Potential of Concentric Cylindrical Insulator

[hime]

Homework Statement

An infinitely long solid insulating cylinder of radius a = 3.2 cm is positioned with its symmetry axis along the z-axis as shown. The cylinder is uniformly charged with a charge density ρ = 22 μC/m3. Concentric with the cylinder is a cylindrical conducting shell of inner radius b = 15 cm, and outer radius c = 20 cm. The conducting shell has a linear charge density λ = -0.41μC/m.

A) What is V(c) - V(a), the potentital difference between the outer surface of the conductor and the outer surface of the insulator?

B)Defining the zero of potential to be along the z-axis (x = y = 0), what is the sign of the potential at the surface of the insulator?

Homework Equations

E.dA=qenclosed/eo=2klambdatotal/r
V=-integral(E.dr)=2klambdatotal*ln(r)
lambda cylinder=7e-8C/m
lambda shell= -4.1e-6C/m
lambda total=lambdacylinder+lambdashell= -3.39e-7

The Attempt at a Solution

for part a: Vc-Va= 2klambdatotal*(ln(c)-ln(a)) = -8 056.39263

for part b: V(a) would be greater than zero because the there would be charge .41e-6 C/m accumulating on the outer surface of the insulating shell.

is the right way to do it? please help!

 

[kuruman]

The Attempt at a Solution

for part a: Vc-Va= 2klambdatotal*(ln(c)-ln(a)) = -8 056.39263

for part b: V(a) would be greater than zero because the there would be charge .41e-6 C/m accumulating on the outer surface of the insulating shell.

is the right way to do it? please help!

Both parts of the attempt at a solution appear to be incorrect.

In part (a), the outer cylindrical conducting shell is an equipotential. The potential difference at r = b is the same as at r = c. It suffices to use Gauss's law to find the electric field in-between conductors and then do the integral, $$V_c-V_a=V_b-V_a=-\int_a^b E~ dr=-\int_a^b \frac{\rho a^2}{2\epsilon_0 r}~ dr= \frac{\rho a^2}{2\epsilon_0 }\ln\left(\frac{a}{b}\right).$$In part (b) the potential at $r=a$ is less than zero because the inner cylinder is positively charged. This means that the electric field lines are radially outward and we know that electric field lines point from higher potential to lower potential.

Frankly, I don't see what the outer conducting shell has to do with anything. Maybe it's a red herring or maybe there are more parts to this problem.

 

[Delta2]

Frankly, I don't see what the outer conducting shell has to do with anything. Maybe it's a red herring or maybe there are more parts to this problem.

Yes either it is a big red herring, aiming to make the student understand that it doesn't affect the E-field in the in-between space, or there are more parts in this problem.

Why its charge density is given in C/m btw shouldn't it be C/m^2? It is a shell with thickness, therefore a 3D structure but we know that because it is conducting all the charge is distributed on the outer surface (or possible in inner surface too).

 

[kuruman]

Why its charge density is given in C/m btw shouldn't it be C/m^2? It is a shell with thickness, therefore a 3D structure but we know that because it is conducting all the charge is distributed on the outer surface (or possible in inner surface too).

Maybe to catch the unsuspecting student. This is a linear charge density $\lambda$ such that the charge enclosed by a cylinder of length $L$ would be $q=\lambda L$. If it were a surface charge density, then $q=\sigma(2 \pi a L)$ Setting the two equal gives you what $\lambda$ ought to be given $\sigma$ or the other way around.

 

[Delta2]

such that the charge enclosed by a cylinder of length

The charge enclosed? But the charge isn't enclosed by the shell, it is on its surface. A hypothetical linear charge density at the shell axis is not equivalent to a surface charge density on its surface because the first produces an E-field in the in-between space...

 

[Orodruin]

The charge enclosed? But the charge isn't enclosed by the shell, it is on its surface. A hypothetical linear charge density at the shell axis is not equivalent to a surface charge density on its surface because the first produces an E-field in the in-between space...

Since the problem clearly states “linear charge density” I think it is pretty clear that this implies charge per length of the conductor along the symmetry axis. That this will be distributed as a surface charge is clear as well. I don’t think this is more confusing than, for example, specifying a spherical conductor with a particular given charge. Yes, the charge will distribute itself on the surface in that case too, but you can still talk about the total charge of the sphere just as you here can talk about the charge per length of the cylinder.

 

[Delta2]

Since the problem clearly states “linear charge density” I think it is pretty clear that this implies charge per length of the conductor along the symmetry axis. That this will be distributed as a surface charge is clear as well. I don’t think this is more confusing than, for example, specifying a spherical conductor with a particular given charge. Yes, the charge will distribute itself on the surface in that case too, but you can still talk about the total charge of the sphere just as you here can talk about the charge per length of the cylinder.

Hmm, but as I said this linear charge density is not equivalent from a physical point of view (meaning that it won't generate the same E-field) to the surface charge density right? It has more of a mathematical meaning regarding the total charge of a cylindrical shell of height L is $q=\lambda L$ but this charge will actually be distributed as surface charge density $\sigma=\frac{q}{2\pi aL}=\frac{\lambda}{2\pi a}$

 

[Delta2]

That this will be distributed as a surface charge is clear as well.

This wasn't so clear to me but it might be clear to you and @kuruman that maybe have a bit more experience of this kind of problems. When I read linear charge density I imagine the charge is distributed along a thin line.

 

[Orodruin]

Hmm, but as I said this linear charge density is not equivalent from a physical point of view (meaning that it won't generate the same E-field) to the surface charge density right? It has more of a mathematical meaning regarding the total charge of a cylindrical shell of height L is $q=\lambda L$ but this charge will actually be distributed as surface charge density $\sigma=\frac{q}{2\pi aL}=\frac{\lambda}{2\pi a}$

You are misunderstanding what I said above. A given linear charge density does not imply that the charge is distributed as a line charge just as a given charge does not imply that it is distributed as a point charge.

 

[Orodruin]

Frankly, I don't see what the outer conducting shell has to do with anything. Maybe it's a red herring or maybe there are more parts to this problem.

You used the fact that it is a conductor to conclude $V_c = V_b$ so I would say it is relevant.

 

[Delta2]

You are misunderstanding what I said above. A given linear charge density does not imply that the charge is distributed as a line charge just as a given charge does not imply that it is distributed as a point charge.

Ok I see now, thanks for clearing this to me!

 

[Delta2]

You used the fact that it is a conductor to conclude $V_c = V_b$ so I would say it is relevant.

Yes right, but that given linear charge density of which we have spoken so much doesn't seem to be of use anywhere for the answer to a) or b). Maybe the problem has more sub questions where it is used.

 

[vanhees71]

Of course, the extra charge on the conductor will spread over the surface at $r=c$ ($r$ being the cylinder coordinate, i.e., $r=\sqrt{x_1^2+x_2^2}$, in addition to the induced charge density by the due to the electric field of the charge of the inner cylinder.

By symmetry the potential is $\Phi=\Phi(r)$ everywhere. For $r<a$ you have
$$\Delta \Phi=\frac{1}{r} [r \Phi'(r)]'=-\rho/\epsilon.$$
Integration gives
$$\Phi(r)=-\frac{\rho r^2}{4 \epsilon} + C_1 +C_2 \ln(r/a) \quad \text{for} \quad r<a.$$
Since there's no singularity at $r=0$, we must have $C_2=0$, and since $\Phi(0)=0$, also $C_1=0$.
The electric field is
$$\vec{E}=E_r \vec{e}_r=-\vec{e}_r \partial_r \Phi, \quad E_r=\frac{\rho r}{2 \epsilon}.$$
For $a<r<b$ you have
$$\Delta \Phi =\frac{1}{r} [r \Phi'(r)]'=0 \; \Rightarrow \; \Phi(r)=C_1'+C_2' \ln(r/a).$$
The potential as well as $E_r$ must be continuous at $r=a$, because there's no surface charge on the inner cylinder, i.e.,
$$-\frac{\rho a^2}{4 \epsilon}=C_1', \quad \frac{\rho a}{2 \epsilon}=-\frac{C_2'}{a},$$
i.e.,
$$\Phi(r)=-\frac{\rho a^2}{4 \epsilon} - \frac{\rho a^2}{2 \epsilon} \ln(r/a), \quad E_r = -\Phi'(r)=\frac{\rho a^2}{2 \epsilon r} \quad \text{for} \quad a<r<b.$$
Then since we have a conductor for $b<r<c$ we must have $\vec{E}=0$ inside, and thus
$$\Phi(r)=-\frac{\rho a^2}{4 \epsilon} - \frac{\rho a^2}{2 \epsilon} \ln(b/a), \quad E_r=0 \quad \text{for} \quad b<r<c.$$
For $r>c$ we have again
$$\Phi(r)=C_1''+C_2'' \ln (r/c).$$
Continuity of $\Phi$ gives
$$C_1''=-\frac{\rho a^2}{4 \epsilon} - \frac{\rho a^2}{2 \epsilon} \ln(b/a).$$
The electric field must have a jump corresponding to the additional surface charge $\sigma_{\text{ext}}=\lambda/(2 \pi c)$ and the induced surface charge $\sigma_{\text{ind}}=\frac{\rho a^2}{2 c}$
$$E_r(c+0^+)=\frac{\lambda+\rho \pi a^2}{2 \epsilon_0 c}=-\frac{C_2''}{c},$$
i.e.,
$$\Phi(r)=-\frac{\rho a^2}{4 \epsilon} - \frac{\rho a^2}{2 \epsilon} \ln(b/a)-\frac{\lambda+\rho \pi a^2}{2 \pi \epsilon_0} \ln(r/c), \quad E_r=-\Phi'(r)=\frac{\lambda + \rho \pi a^2}{2 \pi \epsilon_0r}.$$

 

[Delta2]

@vanhees71 Can you please explain how do you calculate the induced surface charge $\sigma_{ind}=\frac{\rho a^2}{2c}$?

 

[vanhees71]

Take a cylinder with radius $R>c$ in the integral form of Gauss's Law. Then you must get the total charge inside, which is $\rho \pi a^2 L$ from the charge in the inner cylinder and the charge $\lambda L$ from the additional surface charge $\lambda/(2\pi c)$ on the outer surface of the conductor at $r=c$. The surface integral thus is $2 \pi R L E_r(R)=-2 \pi L C_2''=(\rho \pi a^2 + \lambda)L/\epsilon_0$. From this you get $C_2''=-(\rho \pi a^2+\lambda)/(2 \pi \epsilon_0)$.

 

[kuruman]

The charge enclosed? But the charge isn't enclosed by the shell, it is on its surface. A hypothetical linear charge density at the shell axis is not equivalent to a surface charge density on its surface because the first produces an E-field in the in-between space...

@vanhees71 said in more detail what I was going to say after I had a good night's rest. My point was going to be that the volume charge density $\rho$ of the inner cylinder can be recast as a linear charge density $\lambda_{\text{inner}} = \rho \pi a^2.$ That makes it easier to combine with the given outer linear charge density to find the electric field in the region $r>c$ and the (linear) surface charge densities at $r=b$ and $r=c.$

For example, the induced charge per unit length at the inner surface of the outer cylinder must be equal and opposite to the charge per unit length on the inner cylinder, $$\lambda_{\text{outer,b}}=-\lambda_{\text{inner}}= -\rho \pi a^2.$$ This can be converted to a surface charge density by dividing by $2\pi b$, $\sigma_{\text{outer,b}}=-\dfrac{\rho a^2}{2b}.$

 

[Delta2]

Ehm, so the induced surface charge is on the inner surface (r=b) of the cylindrical shell right?

 

[kuruman]

Ehm, so the induced surface charge is on the inner surface (r=b) of the cylindrical shell right?

There will be charge on both surfaces of the outer cylinder such that their sum per unit length is what was put on the outer cylinder, in this case λ = -0.41μC/m.

 

[vanhees71]

At $r=b$ you must have a surface charge such that there's $\vec{E}=0$ between $b<r<c$, i.e., $\sigma_b=-\pi \rho a^2/(2 \pi b)=-\rho a^2/b$. On the outer surface at $r=c$ you have the corresponding opposite induced total charge and the additional charge as described in the problem, i.e., for the surface-charge density
$$\sigma_c=\rho a^2/(2c)+\lambda/(2 \pi c)=(\rho \pi a^2 + \lambda)/(2 \pi c).$$

 

[Orodruin]

It should also be mentioned that, electromagnetism being linear, the solution is the superposition of the case with an overall uncharged cylindrical shell (but with the central homogeneous charge) and the case of a charged cylindrical conductor with no charge inside.

 

[Delta2]

Ok sorry novice question coming up, why on the inner and outer surfaces of the conducting shell we must have equal and opposite induced charges?

 

[Delta2]

Nvm got it, total induced charge must be zero, conservation of charge that is, net charge can't magically appear because of induction.

 

[Orodruin]

Ok sorry novice question coming up, why on the inner and outer surfaces of the conducting shell we must have equal and opposite induced charges?

Nvm got it, total induced charge must be zero, conservation of charge that is, net charge can't magically appear because of induction.

This depends on what restrictions you make. If you specify the charge on the conductor, then the total charge needs to add up to that charge. If instead you specify the potential at the conductor (eg, by grounding it), you can get different charges.

 

[vanhees71]

The induced surface charge at $r=b$ must compensate the charge in the inner cylinder ($r<a$), so that $\vec{E}=0$ in the conductor vanishes. Outside the conductor the electric field must again give the total charge contained in a cylinder with $r>c$, which is the charge in the inner cylinder plus the additional charge on the conductor, which must be on the outer surface to meet all these conditions.