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## 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 E

_{0}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 E

_{0}, 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:

[tex] E=\frac{σ_b}{2πε_0r} [/tex]

And so:

[tex]V_b=\int_c^b \frac{σ_b}{2πε_0r}\mathrm{d}r=\frac{σ_b}{2πε_0} \ln (\frac{b}{c})[/tex]

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:

[tex]\frac{2πblσ_b}{ε_0}=2πblE_0[/tex]

which of course simplifies to [itex]σ_b=ε_0E_0[/itex] and so [itex]σ_c=\text{-}ε_0E_0[/itex], which I also know to be incorrect.

Where am I going wrong on these?