Concentric infinite conducting cylindrical shells, outer one grounded

[CaptainP]

Homework Statement

The figure (found here) shows a cross-sectional view of two concentric, infinite length, conducting cylindrical shells. The inner shell has as an inner radius of a and an outer radius of b. The electric field just outside the inner shell has magnitude E₀ and points radially outward as shown. The grounded, outer shell has an inner radius c and an outer radius of d.

Some of the cylindrical surfaces may be charged. Let σ_a, σ_b, σ_c and σ_d be the surface charge densities on the surfaces with radii a, b, c and d. There are no other charges or conductors in the problem.

Find
(1) an expression for the potential, V, of the inner conducting shell in terms of the given variables and relevant constants, and
(2) an expression for σ_c in terms of E₀, the given radii, and relevant constants.

Homework Equations

ΔV=-∫E⋅dl and Gauss's Law

The Attempt at a Solution

For (1):
From Gauss's law, I know that σ_a=0, and that σ_b>0. Also because the outer cylinder is grounded, the potential at r=d and hence r=c is 0. From Gauss's law:
$$E=\frac{σ_b}{2πε_0r}$$
And so:
$$V_b=\int_c^b \frac{σ_b}{2πε_0r}\mathrm{d}r=\frac{σ_b}{2πε_0} \ln (\frac{b}{c})$$
which I know to be incorrect.

For (2):
Because the electric field in a conductor is necessarily 0, by Gauss's law we know that σ_c=-σ_b. Applying Gauss's law with a Gaussian surface of a cylinder with radius b and length l, we have:
$$\frac{2πblσ_b}{ε_0}=2πblE_0$$
which of course simplifies to $σ_b=ε_0E_0$ and so $σ_c=\text{-}ε_0E_0$, which I also know to be incorrect.

Where am I going wrong on these?

 

[haruspex]

$$E=\frac{σ_b}{2πε_0r}$$

That would be the field due to σ_b, but what about σ_c and σ_d?

 

[CaptainP]

Well to find the E-field from b<r<c, you use a Gaussian surface of a cylinder with radius b<r<c, so the enclosed charge is simply the charge σ_b. The electric fields from σ_c and σ_d should be irrelevant for this part. On top of that, the outer shell is grounded. If it weren't, σ_c would still equal -σ_b so that the E-field in the outer shell is zero, but then σ_d would equal σ_b to keep the outer shell electrically neutral. However, grounding the outer shell allows excess electrons to neutralize the positive charge σ_d, so σ_d is zero; it contributes nothing to electric field.

 

[haruspex]

$$E=\frac{σ_b}{2πε_0r}$$

Doesn't that lead directly to an answer for (2) by setting r = b, E = E₀?

$$V_b=\int_c^b \frac{σ_b}{2πε_0r}\mathrm{d}r=\frac{σ_b}{2πε_0} \ln (\frac{b}{c})$$
which I know to be incorrect.

Is it just the sign that's wrong? (I believe it is wrong, and it happens because you're taking dr and E in opposite directions.)

 

[CaptainP]

Sorry, I should've included the correct answers in the initial post. The correct answer for (1) is $V=\frac{σ_b}{ε_0}b\ln (\frac{c}{b})$. The correct answer for (2) is $σ_c=-ε_0E_0\frac{b}{c}$.

 

[haruspex]

nd hence r=c is 0. From Gauss's law:
$$E=\frac{σ_b}{2πε_0r}$$

I now think that's wrong. σ_b is a charge density. The charge around a circular band width dl is 2πσ_bdl. So the above should read
$$E=\frac{b σ_b}{ε_0r}$$