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enter image description hereIf you put a resistor and a capacitor in series with a 9V battery so that the resistor is in the wire going out from the positive terminal of the battery to a plate of the capacitor.

In my opinion the voltage drop accross the resistor (i m talking about the first miliseconds) would be ONLY the difference between the positive terminal potential and the capacitor's plate connected to the same wire potential.

To illustrate what i mean if The battery's terminals were both +4.5V And -4.5V The voltage drop across the resistor would be only 4.5V(decreases down to 0 as capacitor charges)

My question is this : If in a wire surface charges transfer the electric field accross the wire, what physical process transfers electric field information between the capacitor plates so that the resistor has the expected voltage drop of 9V Thanks,

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  • $\begingroup$ You're viewing a capacitor as two independent isolated plates, but they are not independent nor isolated, a capacitor is a system what works as math rules. There is a displacement current $\endgroup$ – FGSUZ Nov 7 '19 at 23:10
  • $\begingroup$ If you understand the circuit with 9V and 0V at the battery's terminals, then you understand it for +4.5V and -4.5V. $\endgroup$ – JEB Nov 7 '19 at 23:22
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To illustrate what i mean if The battery's terminals were both +4.5V And -4.5V The voltage drop across the resistor would be only 4.5V(decreases down to 0 as capacitor charges)

No, the voltage drop across the wires is always zero, and since initially the capacitor is uncharged so the voltage drop across the capacitor is also 0. So the top of the resistor is 4.5 - 0 = 4.5 V and the bottom of the resistor is -4.5 - 0 - 0 = -4.5 V. So all of the 9 V from the battery will be across the resistor.

what physical process transfers electric field information between the capacitor plates so that the resistor has the expected voltage drop of 9V

The physical process is simply Gauss’ law. As charges accumulate on one plate then per Gauss’ law there is an E field between the plates. This causes an equal and opposite charge to accumulate on the other plate. At the beginning the charge is 0 so the E field is 0 and so the two ends are at the same voltage, -4.5 V.

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  • $\begingroup$ I know that it should be a 9V Voltage drop across the resistor the question is how is it possible. I know this is wrong but from what i understood of capacitors. There would be a 4.5V drop across the resistor and a 4.5V voltage drop across the other wire. $\endgroup$ – mohamed azaiez Nov 8 '19 at 11:29
  • $\begingroup$ The voltage drop across a wire is always 0. $\endgroup$ – Dale Nov 8 '19 at 11:42
  • $\begingroup$ Yes i know that it's wrong but you got what i mean that the voltage drop across the resistor should be only the difference between the positive plate's potential and the positive battery's plate potential not the over all potential difference $\endgroup$ – mohamed azaiez Nov 8 '19 at 11:57
  • $\begingroup$ I think I see your question better now. I have updated my answer, see if that helps. $\endgroup$ – Dale Nov 8 '19 at 12:21
  • $\begingroup$ Thank you, That helps a lot But how are both the capacitor's plate charged at -4.5 V Isn't a capacitor initially neutral and starts gaining charges until it reaches the same potential that the battery's have (4.5V for the positive plate and -4.5V for the negative plate) $\endgroup$ – mohamed azaiez Nov 8 '19 at 12:59

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