|Jul2-12, 01:10 PM||#1|
electric circuit -- Why is the switch needed in this capacitor circuit?
I have two questions with respect to the picture in the attachment.:
I have read, that the capacitor become charged, when the switch is closed. All right, but why the capacitor doesn't became charged before the switch is closed?
At least the left conductor is connected with the positive pole of the battery, so I thought the electrons will leave the left conductor.
In the picture we can see a potential difference V between two points of the electric circuit. Is the potential V always the same between two points, whereas one Point is between the left conductor and the "+"Pole and the other one is between the right conductor and the "-"Pole?
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|Jul2-12, 03:35 PM||#2|
Hi Sogan! Welcome to PF!
a switch is just a capacitor with a very small plate area (A) and a very large separation (d)
since capacitance is 8.85 10-12 A/d, that means that the capacitance of a switch is extremely small (well under 10-15 F)!`
(actually the A/d formula really only applies if A is much larger than d2, but you get the idea)
capacitance is the ability to separate charge, and a switch can separate hardly any charge
if a switch and an ordinary capacitor are in series, the charge across both must be the same
for all practical purposes, the charge (across both the capacitor and the switch) can be considered zero
why can't the battery "pump" electrons onto the nearer plate of the capacitor?
the only difference would be the (extremly small) voltage drop across the extra bit of wire
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