How Does Doubling Capacitance Affect Voltage Distribution in a Series Circuit?

[ortin]

1. Three equal capacitors, all with values C, are in series with a battery of V volts. If the middle capacitor is changed to 2C, what will be the resultant potential across each capicitor? Express your answer in terms of V.

Homework Equations

C = Q / V

The Attempt at a Solution

Initially, when the capacitors are equal, the total voltage will be distrubuted evenly over the capacitors (Kirchhoffs law). Therefore the potential over each = V / 3

If one of the capacitors value is doubled, then following the relation V = Q /C, the potential diffence of the middle capacitor will be halved.

End of attempt

However the next question goes on to discuss the affect of this on the charge per plate. By rearrangement - Q = VC, so if the capacitance is doubled, so should the charge. But then this contradicts the value for V previoulsy found?

Many Thanks!

 

[Shooting Star]

The PEs across each of the capacitors have changed, in such a way that the sum is V. The new PE across 1 and 3 is the same, say V1 (by symmetry). The new PE across the middle one is now V2.

V=2V1+V2.

Suppose C was the capacitance of each at first. The middle one is 2C now.

Now, the charge in the middle one must be equal to the charge of 1st. Using this, what do you get?

 

[ogb p]

Your attempt is correct in terms of the potential difference across each capacitor. When the middle capacitor is changed to 2C, the potential difference across it will be V/2. However, the total charge on the capacitors will still be the same, as charge is conserved in a circuit. This means that the charge on each capacitor will also be halved, since Q = VC.

So, in summary, the potential difference across each capacitor will be V/3 for the two 1C capacitors and V/2 for the 2C capacitor. And the charge on each capacitor will be V/3C for the two 1C capacitors and V/2C for the 2C capacitor. This is consistent with the relation Q = VC.