Science > Physics > Electrostatic forces and principle of virtual work
| Topic: |
Science > Physics |
| User: |
"Marko" |
| Date: |
04 Feb 2005 03:51:12 AM |
| Object: |
Electrostatic forces and principle of virtual work |
When calculating forces on capacitors (parallel plate) when can assume
constant charge on the plates or constant potential at the plates.
Constant charge (isolated system)
Using the principle of virtual work.
External work done from outside is zero, since system is isolated
dW_extrn = 0
For capacitor we have the expression for electrostatic energy
W_e = 1/2*QV
If one plate is moved a infinitesimal distance dx, work is done
dA = F_x*dx
Virtual work principle gives us:
dW_extrn = dW_e + dA
0 = dW_e + dA
dA = F_x*dx = -dW_e
F_x = -d/dx(W_e)
For the parallel plate condensor we get:
W_e = 1/2*QV
dW_e = 1/2*Q*dV
V = E*x = Q/(e0*S)*x
dV = Q/(e0*S)*dx
dA = F_x*dx = dW_e = 1/2*Q*dV = 1/2*Q*Q/(e0*S)dx
F_x = -1/2*Q^2/(e0*S)
When we keep the plates at constant potential (connected to a battery)
I don't understand then how the battery should do work V*dQ
I don't get it say for parallel plate condensor, we have following
Q = CV
where C = e0*S/d
increasing the separation distance between plates capacitanse
decreases.
V is held constant then Q should also decrease, doesn't it means that
the charge from the plates should go to source?
How can the source do work V*dQ, when the separation distance is
increased?
Quoting for the Feynman lectures on Physics vol 2.
"Suppose we had imagined that the condenser was held at a constant
potential difference as we made the virtual displacement. Then we
should have taken
U = 1/2*C*V^2
and in place of Eq (8.15) we should have had Fdz = 1/2V^2dC
which gives a force equal in magnitude to tone in Eq (8.15) (because V
= Q/C) but with opposite sign! Surely the force between the condenser
plates doesn't reverse in sign as we disconnect it for the charging
source. Also, we know that two plates with opposite electrical charges
must attract. The principle of virtual work has been incorrectly
applied in the second case - we have not taken into account the
virtual work done on the charging source. That is, to keep the
potential constant at V as the capacity changes, a charge VdC must be
supplied by a source of charge. But this charge is supplied at a
potential V, so the work done by the electrical system which keeps the
potential constant is V^2dC. The mechanical work Fdz plus this
electrical work V^2dC together make up the change in the total energy
1/2V^2dC of the condenser. Therefore Fdz is -1/2V^2dC, as before"
Why is source delivering the charge VdC??
Thanks for you reply
/Mario
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| User: "John C. Polasek" |
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| Title: Re: Electrostatic forces and principle of virtual work |
04 Feb 2005 09:58:01 PM |
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On 4 Feb 2005 01:51:12 -0800, (Marko) wrote:
When calculating forces on capacitors (parallel plate) when can assume
constant charge on the plates or constant potential at the plates.
Constant charge (isolated system)
Using the principle of virtual work.
External work done from outside is zero, since system is isolated
dW_extrn = 0
For capacitor we have the expression for electrostatic energy
W_e = 1/2*QV
Change the equation to incorporate the constant charge.
Since V = Q/C, you get W = Q^2/2C
C = epA/x
W = Q^2*x/2Aeps
F = -dW/dx = - Q^2/2Aeps
snip
/Mario
John Polasek
If you have something to say write an equation.
If you have nothing to say, write an essay.
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