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Capacitance in a conducting cylinder

  • Thread starter oreosama
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  • #1
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Homework Statement



A cylindrical capacitor of length L consists of a solid conducting core with a radius R and an outer outer hollow conducting tube with an inner radius 3R. A voltage [tex]V_{ab}[/tex] is applied between the two cylinders. Assume L >> R which means we can neglect edge effects.

Given [L, R, [tex]V_{ab}[/tex], K]

Determine:

The charge per unit length for the capicitor
The voltage at 2R
The total charge on the capacitor
The electric field at 2R
the capacitance
the energy stored in the capacitor

Find all answers again if a dielectric K is inserted between the cylinders

Homework Equations



[tex]\lambda = \frac{q}{L}[/tex]
[tex]\oint E \cdot dA = \frac{q}{\epsilon_0}[/tex]
[tex]V = \int E \cdot dl[/tex]
[tex]C = \frac{q}{V}[/tex]

The Attempt at a Solution



The question is kind of all over the place and it makes me have doubts on what im doin

[tex]\lambda = \frac{q}{L}[/tex] but we need to solve for q at some point

[tex]V(2R) = \frac{\lambda}{2 \pi \epsilon_0} \cdot \ln{R}[/tex] (use gauss law to solve for E apply to V formula with 3R as "0")

I'm not sure what direction to approach finding the charge(or if im doing the rest so far right). capacitance isnt given so C = q/v wont help.
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement



A cylindrical capacitor of length L consists of a solid conducting core with a radius R and an outer outer hollow conducting tube with an inner radius 3R. A voltage [tex]V_{ab}[/tex] is applied between the two cylinders. Assume L >> R which means we can neglect edge effects.

Given [L, R, [tex]V_{ab}[/tex], K]

Determine:

The charge per unit length for the capicitor
The voltage at 2R
The total charge on the capacitor
The electric field at 2R
the capacitance
the energy stored in the capacitor

Find all answers again if a dielectric K is inserted between the cylinders

Homework Equations



[tex]\lambda = \frac{q}{L}[/tex]
[tex]\oint E \cdot dA = \frac{q}{\epsilon_0}[/tex]
[tex]V = \int E \cdot dl[/tex]
[tex]C = \frac{q}{V}[/tex]

The Attempt at a Solution



The question is kind of all over the place and it makes me have doubts on what im doin

[tex]\lambda = \frac{q}{L}[/tex] but we need to solve for q at some point

[tex]V(2R) = \frac{\lambda}{2 \pi \epsilon_0} \cdot \ln{R}[/tex] (use gauss law to solve for E apply to V formula with 3R as "0")

I'm not sure what direction to approach finding the charge(or if im doing the rest so far right). capacitance isnt given so C = q/v wont help.
Your formulas for V is not correct. Check the units: The argument of he logarithm has to be dimensionless.
You have definite integral when calculating the potential difference between the inner and outward cylinders. Also, check the sign in the firs equation in red.
At the end, substitute λ=q/L, then you can get the capacitance as C=q/V.

ehild
 
  • #3
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oh, whoops. I meant ln(3R/2R) -> ln(3/2). you're saying use V(2R) as basis for the capacitance? I was kind of confused on how V(2R) contains all the charge we're talking about, but since the voltage I measured is from the surface of both conductor surfaces it should catch it all? I'm trying to conceptually understand :)
 
  • #4
ehild
Homework Helper
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You have integrate the electric field between the cylinders to get the potential difference between them.
It is [tex]\Delta V = V(3R)-V(R) = - \int_R^{3R}{Edr}[/tex] . There are equal charges on both cylinders, with opposite signs; q and -q. C=q/|ΔV|.

ehild
 
Last edited:
  • #5
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The concept of voltage being constant in a capacitor for some reason was lost on me.. that makes sense. this voltage is the same as the given correct? so

[tex]q = \frac{VL 2 \pi \epsilon_0}{\ln{1/3}}[/tex]

[tex]C = \frac{L 2 \pi \epsilon_0}{\ln{1/3}}[/tex]
 
  • #6
ehild
Homework Helper
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Yes, only the sign has to be changed to positive. The voltage is equal to the potential drop between the surfaces, it is -ΔV. If you replace ln(1/3) with ln(3) your result is correct.

ehild
 
Last edited:
  • #7
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Oh my... Does this apply to everything I've done so far? Are you just saying "voltage" is the "opposite of the potential difference"?
 
  • #8
ehild
Homework Helper
15,478
1,854
There is some confusion when using the word potential difference and voltage. Some people use voltage for potential, or potential difference. In Ohm's Law or in the definition of capacitance V=IR or V=q/C, V stands for the voltage a voltmeter would read if you connect its positive terminal to the the positive side of the resistor/capacitor and the negative terminal to the negative side.
The electric field is equal to the negative gradient of the potential, the integral of E from 1 to 2 gives -(V2-V1). If you move in the direction of E, the integral is positive, and the potential drops from 1 to 2. V2 -V1 is called potential difference, but you use the positive V1-V2 when calculating capacitance.

Rather confusing, I know...

ehild
 

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