Force between cylindrical capacitors

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unscientific
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Homework Statement



Typed problem 10 on a capacitor: two coaxial open-ended conducting cylinders of radii a and b (a < b) and length l, derive the capacitance; then force between the cylinders for parts (a) charged and disconnected, and (b) connected to a voltage source V.


Homework Equations





The Attempt at a Solution



Handwritten derivation of a coaxial cylindrical capacitor: C = 2πε0 L / ln(b/a), V = Q/(2πε0 L) ln(b/a), and energy U = ½CV² = ½QV = ½Q²/C with dU = -½V² dC and F = ∂U/∂x. Small sketch of a cylinder with inner radius a and outer radius b.


I'm not sure what is being varied here, radial separation between capacitors x = (b-a) or whether the capacitors can slide up and down so as to change the length L of the capacitor..
 
unscientific said:
I'm not sure what is being varied here, radial separation between capacitors x = (b-a) or whether the capacitors can slide up and down so as to change the length L of the capacitor..

Since the question states that the inner cylinder is partially withdrawn along the common axis, I think it's L that is varied.
 
TSny said:
Since the question states that the inner cylinder is partially withdrawn along the common axis, I think it's L that is varied.

OK, so for constant Q, the force is (1/2)Qo(∂V/∂L), and constant V, the force is (0.5)V^2 (∂C/∂L):

Handwritten derivation under the heading Constant Q: F = (1/2) Q dV/dx, then = Q^2 ln(b/a) / (4 pi epsilon_0 Q^2), then = V^2 pi epsilon_0 / ln(b/a).

Handwritten derivation headed 'Constant V:' giving dU = (1/2)V^2 dC, then F = (1/2)V^2 dC/dL, then F = (1/2)V^2 (2 pi epsilon_a / ln(b/a)).
 
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How are you defining x? Is it the amount by which the inner conductor is withdrawn or the amount by which the inner conductor is still inside the outer conductor? Or does x represent something else?

How would you write C in terms of x?
 
Won't it be much better to use the equation ##U=\frac{Q^2}{2C}## and calculate the force when the charge is constant?
 
TSny said:
How are you defining x? Is it the amount by which the inner conductor is withdrawn or the amount by which the inner conductor is still inside the outer conductor? Or does x represent something else?

How would you write C in terms of x?

Sorry i meant L instead of x.
 
Pranav-Arora said:
Won't it be much better to use the equation ##U=\frac{Q^2}{2C}## and calculate the force when the charge is constant?

The C is in the denominator, which gives a C-2(∂C/∂L) term