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If we charge a cap and then pull it apart so that voltage increases, how does charge stay the same now the capacitance is smaller?
Assuming capacitance drops below the point to hold the charge.
Tnx for the effort nyphdinmd but it doesn't really answer my question. I am well aware of formulas for capacitance. The answer to my question would be that decrease of capacitance due to pulling cap apart decreases it's ability to be charged (due to weakened field lines between the plates), but since it is ALREADY CHARGED, pulling cap apart does not and cannot change the amount of electrons stored on the negative terminal.
2 Answers
- nyphdinmdLv 73 years agoFavorite Answer
For a parallel plate capacitor, the capacitance, C is given by: C = e0*K*A/0where e0 = permittivity of free space, K = dielectric constant of media between the plates, A = area of the plates and z0 = distance between the plates.
Now lets say we charge the capacitor to some voltage V0. Then by C = Q/V we can find an expression for Q
Q = CV0 = e0*K*A*V0/z0
If we isolate the capacitor, no charge can enter or leave. So we have Q = const = V(z)*e0*K*A/z where V is a function of plate position and V(z0) = V0. Thus
V(z) = z*Q/(e0*K*A)
So as z gets larger (plates get pulled apart) V increases. This reflects the facgt that you need to do work to pull the plates apart and the work increases the potential energy between the plates (conservation of energy).
