I'm a bit confused as to why there is work done against the electric field when charging a capacitor, as charge is moving from high potential to low potential.

For the sake of this question, I'll assume conventional current. Thus, there's an electric field flowing from the anode to what will become the positively charged conducting parallel plate. Let's say the battery is 12V. At first, when the battery is connected to the plates, the voltage of the plates would be zero. So positive charges would want to flow downstream the electric field. Then, the voltage difference between the battery and the positive plate decreases until it reaches zero. But I don't see how at any point the battery has to do any work against the electric field to push charges, as positive charges will naturally want to flow from high voltage to low voltage. I've seen the answer relate to force against the electric field between the plates, but I don't see how this is relevant as charges are not moving between the plates.


The man is like the battery,he needs to do work against electric field to accumulate charges on the upper plate because the charges that the plate already has will oppose further accumulation of charges

Source of the image : MIT OpenCourseWare

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  • $\begingroup$ That's a nice and brilliant presentation. In explaining things like this, a picture is worth more than a thousand words. $\endgroup$ – UKH Feb 18 '17 at 7:24
  • $\begingroup$ Thank You ,but I should mention that ,this pic is from Mit open course ware. $\endgroup$ – Paul Feb 18 '17 at 7:34
  • $\begingroup$ Thank you @Asgardian, but let's say the battery is 12V and the capacitor builds up charge until it's at 12V. When it's less than 12V, from what I understand there should not be any work to put another charge on the plate, as a drop in voltage is where positive charges want to go (i.e. they will increase kinetic energy to get there) $\endgroup$ – rb612 Feb 21 '17 at 5:52
  • $\begingroup$ @rb612 : Lets suppose you want to fill a box with electrons and you are putting electrons one by one in the box.Atfirst it won't be a problem,but after sometime the electrons that are already inside the box will apply repulsive force on the electron you want to put inside ,so you need to do some work to put it inside. $\endgroup$ – Paul Feb 21 '17 at 7:17
  • $\begingroup$ @rb612 :Now the battery does something similar ,it creats an electric field to bring electrons from one plate to the other.Atfirst there will be no problem ,but after sometime the electrons on the other plate will apply a repulsive force on the electron the battery is trying to bring so the battery needs to do work. $\endgroup$ – Paul Feb 21 '17 at 7:26

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