Finding electric potential given a cylindrical configuration

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carloz
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Homework Statement



You have two concentric cylinders. The inner cylinder has radius a and the external cylinder has radius b. Find the electric potential in the region between the cylinders.

[Hint: The final equation takes the form V(r) = constant1 - constant2 ln(something) ]


Homework Equations



I think:
Gauss's law: [itex]\Phi[/itex]=EA=q/ε
Coulomb's law: E=q/(4πεr^2)

The Attempt at a Solution



I really don't know.
 
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Okay

So

EA=q/ε

E=q/(εA)

A=2πrL

dA=2πLdr

dE=q/(ε2πLdr)

E = ∫(limits a and b) q/(ε2πLdr) = q/(ε2πL) ∫(limits a and b) 1/dr

Am I on the right track? If so, how to evaluate ∫(1/dr) ?

Thank you.
 
Do you know the answer to this? I am just learning this material also, so I don't know if I am thinking through it correctly. But the outer shell should be irrelevant as the electric field is going to be zero for any point inside it assuming it is infinitely long, correct? For the electric field at r distance from the center of cylinder with radius a, would its magnitude be λ/(2πεr) (derived with Gauss's law)? If so, you would end up with Δv(r) = v(r) (taking v_i = 0) = λln(a)/(2πε) - λln(r)/(2πε) where a <= r <= b, λ = q/l. But I have no idea if that is correct.
 
carloz said:
Okay

So

EA=q/ε

E=q/(εA)

A=2πrL

dA=2πLdr

dE=q/(ε2πLdr)

E = ∫(limits a and b) q/(ε2πLdr) = q/(ε2πL) ∫(limits a and b) 1/dr

Am I on the right track? If so, how to evaluate ∫(1/dr) ?

Thank you.
No. First off, what's your Gaussian surface and what expression is equal to the amount of charge enclosed in it?