Force between cylindrical capacitors

unscientific
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Homework Statement



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Homework Equations





The Attempt at a Solution



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Since the question states that the inner cylinder is partially withdrawn along the common axis, I think it's L that is varied.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Since the question states that the inner cylinder is partially withdrawn along the common axis, I think it's L that is varied.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

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How are you defining x? Is it the amount by which the inner conductor is withdrawn or the amount by which the inner conductor is still inside the outer conductor?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How are you defining x? Is it the amount by which the inner conductor is withdrawn or the amount by which the inner conductor is still inside the outer conductor?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
 
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