-2
$\begingroup$

I want to gain a deeper understanding as to how a capacitor actually gets charged.

Consider a situation in which an uncharged parallel plate capacitor is connected to a battery using two wires. I know that charges will keep flowing until the time comes when the potential of each plate equals the potential of the corresponding battery terminal. At this stage the net field in the wire becomes zero, and no more current flows.

  1. Can anyone explain how this actually happens using surface charges in the wires?

  2. Just before charges accumulate on the capacitor plate, do they decelerate? If so, what causes this deceleration? Since we started off with a neutral capacitor, the net field produced by the positive and negative plates at an external point should be zero. So how do the electrons suddenly lose the kinetic energy they initially had? Can this be explained by surface charges as well?

Please correct me wherever I am conceptually wrong.

Thanks in advance

$\endgroup$
  • $\begingroup$ Please give me a detailed answer $\endgroup$ – Newton May 19 '16 at 14:52
  • $\begingroup$ Have you tried to get this information from the internet? What websites have you looked at? $\endgroup$ – sammy gerbil May 19 '16 at 14:59
  • $\begingroup$ (1) Why the emphasis on surface charge? The source of the electric field in the wires is the battery. (2) Why do you think that any given electron has significant kinetic energy such that they need to decelerate? Overall, it would seem you have some conceptual issues with the mechanisms for conductivity in ordinary metals. $\endgroup$ – Jon Custer May 19 '16 at 15:02
  • $\begingroup$ @Jon Custer. I have learnt that the field in a resistor, for example is much higher than the field outside the resistor due to the surface charge accumulation at the boundary, as well as due to the battery , of course. Please correct me if I am wrong. $\endgroup$ – Newton May 19 '16 at 15:06
1
$\begingroup$

To question 2: When the electron reaches the end of a conductor, it would have to move into the air, which is an isolator. The entire conductor is at equal potential, which is much much lower than the potential at a point out in the air. So it reaches the end and stops, since it is only driven by the potential difference

$$F=\frac{dU}{dx}$$

in it's rush to reach a point of lowest potential.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.