Concentric infinite conducting cylindrical shells, outer one grounded

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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 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 σa, σb, σc and σd 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 σc 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 σa=0, and that σb>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 σc=-σb. 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?
 

Answers and Replies

  • #2
haruspex
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[tex] E=\frac{σ_b}{2πε_0r} [/tex]
That would be the field due to σb, but what about σc and σd?
 
  • #3
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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 σb. The electric fields from σc and σd should be irrelevant for this part. On top of that, the outer shell is grounded. If it weren't, σc would still equal -σb so that the E-field in the outer shell is zero, but then σd would equal σb to keep the outer shell electrically neutral. However, grounding the outer shell allows excess electrons to neutralize the positive charge σd, so σd is zero; it contributes nothing to electric field.
 
  • #4
haruspex
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[tex] E=\frac{σ_b}{2πε_0r} [/tex]
Doesn't that lead directly to an answer for (2) by setting r = b, E = E0?
[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.
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.)
 
  • #5
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Sorry, I should've included the correct answers in the initial post. The correct answer for (1) is [itex]V=\frac{σ_b}{ε_0}b\ln (\frac{c}{b})[/itex]. The correct answer for (2) is [itex]σ_c=-ε_0E_0\frac{b}{c}[/itex].
 
  • #6
haruspex
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nd hence r=c is 0. From Gauss's law:
[tex] E=\frac{σ_b}{2πε_0r} [/tex]
I now think that's wrong. σb is a charge density. The charge around a circular band width dl is 2πσbdl. So the above should read
[tex] E=\frac{b σ_b}{ε_0r} [/tex]
 

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