Suppose the plates of a parallel plate capacitor have charges +2Q and -Q initially.

Then if they are connected with a battery of emf V what will be the charge distribution on the plates of capacitor?

I tried it by initially getting the charge distribution. But when the battery is connected i am getting confused.

  • 1
    $\begingroup$ Same as with zero initial charge plus 0.5Q on each plate $\endgroup$
    – LLlAMnYP
    Commented May 25, 2016 at 17:22
  • $\begingroup$ What is capacitance, initially,charge is Q what intial voltage ,it look incomplete to me $\endgroup$ Commented Sep 13, 2019 at 4:10

2 Answers 2


Start off by considering what the charge distribution would be like without the battery being connected. The charge distribution must be such that there is no electric field inside either of the plates.
The consider which of the charges will change when the battery is connected.

  • $\begingroup$ Do we need the value of the capacitance then, also the voltage?@Farcher $\endgroup$
    – Sidarth
    Commented May 30, 2021 at 4:12
  • $\begingroup$ The capacitance is a function of the geometry of the system of plates not the charges on them. $\endgroup$
    – Farcher
    Commented May 30, 2021 at 7:17

Once you connect the capacitor to the battery the voltage across the capacitor must the the same as that of the battery. Once you know the voltage across the capacitor you can calculate the charge distribution.

What is the role of the extra initial charge on the plates? Nothing, because the extra charges will be quickly killed by the holes and the electrons of the battery.


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