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Say, I have two charged capacitors, one 3mF and one 2mF. The voltage across them are 20V and 30V respectively. Now if I connect the two capacitors side by side as shown below, what will be the voltage across each capacitor?

     open -------||----------------||-----------open
             20V, 3mF            30V, 2mF
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  • $\begingroup$ They are connected in series, so the voltage doesn't change. They should still have 20V and 30V respectively. $\endgroup$
    – Lemon
    Aug 8, 2012 at 5:21

4 Answers 4

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Charges on all plates for capacitors in series must be equal. Q=CV for each capacitor. If you look at the charge per plate before you connect them in series, both have the same charge. So, after you connect them, nothing will change (no charge will flow). Since the capacitance of each capacitor doesn't change either, they keep the same voltages.

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  • $\begingroup$ How you are saying that the capacitors are in series? The two extreme terminals are open/ floating. Agree that charge on both the capacitors are same but what about the voltage difference between the rhs plate of 3mF cap and lhs plate of 2mF cap which are connected by a wire? Will there not be any flow of electron due to difference in voltage? $\endgroup$
    – B Biswas
    Aug 8, 2012 at 9:50
  • $\begingroup$ Are you asking the voltage region between the capacitors then? $\endgroup$
    – Lemon
    Aug 8, 2012 at 18:04
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    $\begingroup$ @B Biswas That is exactly what I'm saying. The voltage across the capacitor has nothing to do with the movement of charge already on the capacitor, it tells you what will happen to new charges introduced to the system. As it stands, things are in equilibrium. As to how one can claim the capacitors are in series, the external connection (or lack thereof) doesn't affect anything about claiming that these two capacitors are in series with each other. $\endgroup$
    – Mitchell
    Aug 8, 2012 at 18:48
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There will be no change in the system. Nothing moves. Nothing changes.

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since circuit is open no charge is shared. if the circuit is closed ie capacitors are connected by conducting wire then charge will be shared. charge would flow from higher potential to lower until they come to common potential.

In the given circuit no charge is shared and potential across both the capacitors remains same.

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The charges on the two "outer" plates have nowhere to go and hold the charges on their respective "inner" plates in position, ie not allowing them to move.
As far a the distribution of charges on the two capacitors is concerned it makes no difference as whether the two "inner" plates are connected or not connected.

The potential difference between the two outer plates will be either $30+20 = 50\, \rm V$ or $30 -20 =10\,\rm V$ depending on the sign of the charges on the two inner plates.
Connecting the plates with opposite charges $\Rightarrow 50\,\rm V$ and connecting the plates with like charges $\Rightarrow 10\,\rm V$.

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