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Finding the electric potential difference?

  • Thread starter hanyu
  • Start date
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1. Homework Statement
Please don't do the problem for me, but explain to me in detail each steps I should take and why in my calculations (so I didn't include any numbers).

There's an insulating infinite cylinder of equal charge distribution at origin (density given- rho). Let's say it's radius is a. Then some radius outwards, there's another charged conducting cylindrical shell (given linear charge density- lambda) that encloses the original cylinder with some thickness. It's inner surface is at a distance (radius) b from the origin, and its outer surface c from the origin.

I basically need to find V(c) - V(a), the pot. diff at surface of insulator, to outer surface of the shell.

2. Homework Equations

V(r) = V(r) - V(inf) = INT[E.dl] from inf to r.
E = lamda/ (2pi e0 r)
Area of circle: pi r^2

3. The Attempt at a Solution

So the potential specifically for inf. cylinders should be something like:
V(r) = lamda/ (2pi e0) INT[1/r dr] from inf to r.

Since rho is useless, I'll have to change it into lambda, by multiplying rho by the area of the cross-sectional circle of the cylinder, (pi a^2). Then we can now view the insulating cylinder as a thin wire at the origin (along z-axis) with lambda = rho*pi*a^2.

to find V(c)-V(a), we must do multiple integrals.
for V(c),
INT[E.dl]<from inf to c>

for V(a),
INT[E.dl]<from inf to c> + INT[E.dl]<from c to b> + INT[E.dl]<from b to a>.
INT[E.dl]<from c to b> should be 0 because the E-filed within the conductor is 0.

so V(c)-V(a) = -INT[E.dl]<from b to a>.
= -lamda/ (2pi e0) INT[1/r dr] from b to a, where lambda as we found is rho*pi*a^2.

= -(rho*a^2)/(2 e0) * (ln(a)-ln(b))

...but apparently I'm wrong D: What did I do wrong, how should I have done it and why??
 

Simon Bridge

Science Advisor
Homework Helper
17,823
1,637
From scratch:
1. exploit the symmetry of the situation ... change to cylindrical polar coordinates.
2. use [itex]\epsilon_0 \nabla^2\phi=\rho_{free}[/itex]

But you can finesse it using known results.
1. the field inside a charged conducting surface is zero - so the potential is a constant
2. the field outside a long cylinder of charge is the same as a line of charge through the origin - you should have a solution for that.
3. the superposition principle.
4. the field is the gradient of the potential.

So,
V(a) is that due to a line of charge through the origin + a constant.
V(c) is that due to a superposition of two lines of charge.

Is that what you are trying to do?
 

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