# Adding a capacitor in series question

## Homework Statement

Three capacitors are connected in series across a
DC power source that produces a constant potential
difference. A fourth capacitor is now added in series with
the other three. As a result of adding this capacitor, the
total energy stored in the system of capacitors has? (decreased, increase, no change)
How would this change if the capacitor was added in parallel?

## Homework Equations

C=Q/V E=1/2qV E=1/2q^2/C

## The Attempt at a Solution

So adding a capacitor has decreased the capacitance. I think the potential difference increase and the charge decrease the same amount? not sure

If it were in parallel the voltage would have to be the same but the charge could be different. The capacitance would increase. I'm unsure of how to go about this?
can someone explain what would happen how we know the energy?

the answer is decrease and increase respectively.

Simon Bridge
Homework Helper
You can do this two ways.
1. use the physics ... sketch the arrangement as parallel-plate capacitors and work out the physically equivalent system. i.e. you put two identical caps in parallel, the equivalent 1-cap system has two plates the same distance apart as the two original ones, but twice the surface area. What are the consequences of that?

2. just use the equations you have listed and plug the numbers in.

I guess more of my question is what happens to charge and voltage. I'll do that but it's difficult for me to gain intuition that way without someone explaining it to me.

It seems like if they are added in parallel the charge would increase and the capacitance would increase but again im not positive and maybe the voltage stay constant? I think thats right but i dunno.

Simon Bridge
Homework Helper
What is the relationship between charge on the plates and the voltage between them?

Directly related? Higher voltage higher charge? via qk/r^2*d

Simon Bridge
Homework Helper
For the same capacitor dimensions, the higher the voltage the higher the charge.
If you doubled the separation of the plates, however, keeping the voltage the same, what happens to the charge?
How about if you doubled the area of the plates, keeping the same separation and voltage?

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html

rude man
Homework Helper
Gold Member

## Homework Statement

Three capacitors are connected in series across a
DC power source that produces a constant potential
difference. A fourth capacitor is now added in series with
the other three. As a result of adding this capacitor, the
total energy stored in the system of capacitors has? (decreased, increase, no change)
How would this change if the capacitor was added in parallel?

## Homework Equations

C=Q/V E=1/2qV E=1/2q^2/C

## The Attempt at a Solution

So adding a capacitor has decreased the capacitance. I think the potential difference increase and the charge decrease the same amount? not sure

If it were in parallel the voltage would have to be the same but the charge could be different. The capacitance would increase. I'm unsure of how to go about this?
can someone explain what would happen how we know the energy?

the answer is decrease and increase respectively.

What's the formula for energy in a capacitor of value C with voltage V across it?
Does it matter if C is one capacitor or a bunch in series?
What happens to C when you add a capacitor in series?

It should also answer question 2. What happens to C when you parallel with another capacitor?

If you double the area the capacitance increases by 2 voltage would not change but charge would increase by a factor of 2 I'm guessing now.

If you double the separation while keeping voltage constant the charge would decrease? the only proof of that I can think of is Q=CV V stays the same and C must decrease because the separation was increased.

Here's what I've concluded based off a conceptual exercise in my book.
Parallel addition= no change in voltage but increase in charge(more SA of plates I'm guessing) totally makes sense to me.
Series addition= decreases voltage per capacitor and hence decreases charge per capacitor. I'm guessing because C=Q/V if C decreases then Q and V both have to decrease but I cannot figure out exactly why that is. Taking from Simon's statement "higher voltage higher charge" that would make sense. BUT how does the voltage decrease? that statement does seem to check out.

After thinking about the addition in series how does that make sense isn't the overall voltage increased since the separation total is larger and not too sure what happens to charge still. So confused. I know the charge is constant between the plates but I dont understand what adding does.
l l-l l-l l -> l l
l l-l l-l l-l l -> l l greater potential difference with the addition.

If the voltage stays the same then and we add capacitors the ratio to voltage per capacitor should decrease decreasing the charge as well. But how is the voltage the same after the addition of a capacitor because it will increase the potential difference like above.

CWatters
Homework Helper
Gold Member
Am I missing something here?.....

First you have three capacitors in series with a DC source. Lets assume that has voltage VDC. The capacitors will be charged to VDC.

Then the circuit is opened and another capacitor is inserted in series.

Presumably the voltage on the 4th capacitor is 0V before it's connected otherwise it would bring an unknown energy to the system.

The three original capacitors will still be charged to VDC so there is no voltage drop across the 4th capacitor. No current flow, no change in energy stored in the capacitors.

rude man
Homework Helper
Gold Member
Am I missing something here?.....

First you have three capacitors in series with a DC source. Lets assume that has voltage VDC. The capacitors will be charged to VDC.

Then the circuit is opened and another capacitor is inserted in series.

Presumably the voltage on the 4th capacitor is 0V before it's connected otherwise it would bring an unknown energy to the system.

The three original capacitors will still be charged to VDC so there is no voltage drop across the 4th capacitor. No current flow, no change in energy stored in the capacitors.

You are absolutely right. Good point!!!

Only if the 4th capacitor had initial charge on it would there be current flow i = -V40δ(t) where V40 is the initial charge on the 4th capacitor. The - sign assumes the initial charge polarity of capacitor 4 lines up with the others (+ of C4 connected to + power supply terminal).

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"Three capacitors are connected in series across a
DC power source that produces a constant potential
difference. A fourth capacitor is now added in series with
the other three. As a result of adding this capacitor, the
total energy stored in the system of capacitors has

B. decreased."

Straight from a lecture slide. So you're saying my lecturer is wrong?

gneill
Mentor
"Three capacitors are connected in series across a
DC power source that produces a constant potential
difference. A fourth capacitor is now added in series with
the other three. As a result of adding this capacitor, the
total energy stored in the system of capacitors has

B. decreased."

Straight from a lecture slide. So you're saying my lecturer is wrong?
He's not necessarily wrong, but clearly he has committed an error of omission: he failed to point out a hidden assumption, namely that that all the capacitors be discharged (i.e. the experiment is "reset" to zero) before the battery is reconnected for each energy measurement.

Ah well. Someone help me get closer to understanding the answer?

gneill
Mentor
Making the "unstated assumption" that each experiment is run beginning with uncharged capacitors, then if you know how to add capacitances in series and capacitances in parallel to yield the net equivalent capacitance, and if you further know how to calculate the energy stored on the resulting net capacitance given the potential across it, then you should be able to answer the question without much difficulty...

Sigh....

Again. Can someone explain to what happens to charge and voltage in the scenario I listed.

Alright I've figured it out.

What was tripping me up is I was thinking that the overall voltage must change because the overall separation changed. But if I treat it as a equiv capacitor then the voltage is the same because its one equiv capacitor and the voltage is supplied by the battery and the battery doesn't change. The capacitance decreases and the voltage across the eq capacitor stays the same in both scenarios. Now the voltage across each individual capacitor decreases because the overall voltage supplied stays the same. Less to go around I guess. Charge decreases because I'm as someone else put it "effectively fitting a smaller capacitor on" so it can hold less charge and mathematically speaking Q=CV if V is constant and C decreases then Q must decrease.

rude man
Homework Helper
Gold Member

In the scenario with the (uncharged) 4th capacitor plugged into the series-connected capacitors, nothing happens, as c.w. pointed out. The 4th capacitor remains uncharged while the others retain their original charges and voltages.

But if the 4th capacitor had voltage V4 across it initially, a charge difference of -V4/(1/C1 + 1/C2 + 1/C3 + 1/C4) is imposed on each of the four capacitors. This includes C4 so even C4 will have a different charge on it after hookup than originally.

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Simon Bridge
Homework Helper
"Three capacitors are connected in series across a
DC power source that produces a constant potential
difference. A fourth capacitor is now added in series with
the other three. As a result of adding this capacitor, the
total energy stored in the system of capacitors has

B. decreased."

Straight from a lecture slide. So you're saying my lecturer is wrong?

He's not necessarily wrong, but clearly he has committed an error of omission: he failed to point out a hidden assumption, namely that that all the capacitors be discharged (i.e. the experiment is "reset" to zero) before the battery is reconnected for each energy measurement.

I'd say it is more that he failed to point out that the system reached equilibrium. That would have already been included as part of this stage in the course.

Sigh....

Again. Can someone explain to what happens to charge and voltage in the scenario I listed.

Imagine the capacitors are parallel-plates.
Each has equal area A and separation d.

How does the area and the separation affect the capacitors ability to store energy at a given voltage?

Two such capacitors in series would have an effective area of A and an effective separation of 2d ... sketch it out to see why.

Two in parallel would have area 2A and separation d.

Compare the energy stored in each configuration, at a given voltage, with that for a single capacitor.

What happens to each configuration's ability to store energy at a particular voltage if you added a 4th capacitor in each case?

rude man
Homework Helper
Gold Member
Alright I've figured it out.

What was tripping me up is I was thinking that the overall voltage must change because the overall separation changed. But if I treat it as a equiv capacitor then the voltage is the same because its one equiv capacitor and the voltage is supplied by the battery and the battery doesn't change. The capacitance decreases and the voltage across the eq capacitor stays the same in both scenarios. Now the voltage across each individual capacitor decreases because the overall voltage supplied stays the same. Less to go around I guess. Charge decreases because I'm as someone else put it "effectively fitting a smaller capacitor on" so it can hold less charge and mathematically speaking Q=CV if V is constant and C decreases then Q must decrease.

If by 'figured it out' you mean you went with "decreased", that is wrong.

rude man
Homework Helper
Gold Member
I'd say it is more that he failed to point out that the system reached equilibrium. That would have already been included as part of this stage in the course.

This has nothing to do with 'reaching equilibrium'. The situation is exactly as CWatters described in post 9.

CWatters
Homework Helper
Gold Member
Sorry if my earlier post confused you.

If I understand correctly your lecturer is saying the correct answer is "adding a 4th capacitor in series causes the energy to decrease". That is only correct if the question is badly worded.

Although not stated in the question lets assume all the capacitors (value C) are discharged before each circuit is built...

In your problem there are three circuits:

a) Three capacitors in series
b) Four capacitors in series
c) Three in series with one in parallel.

In circuit a) you have three capacitors in series. The equivalent capacitance is

Ce = 1/(1/C +1/C + 1/C) = C/3

See here for an explanation...

http://www.tutorvista.com/content/p...ential-capacitance/combination-capacitors.php

The energy stored in a capacitor Ce is equal to

0.5 * Ce * VDC2
substituting gives
0.5 * C/3 * VDC2
or
C/6 * VDC2

In circuit b) the maths is the same except you have 4 capacitors...

Ce = C/4

Energy is
0.5 * C/4 * VDC2
or
C/8 * VDC2

So the amount of energy in the 4 capacitor circuit is lower (decreased) compared to the three capacitor circuit (but only if the capacitors are discharged between building each circuit. That's not what the question actually says but might be what your lecturer meant to say).

Now for circuit c)...

The Equivalent capacitance of three in series and one I parallel is

Ce = C/3 + C = 4C/3

The energy stored is..

0.5 * 4C/3 * VDC2
or
2C/3 * VDC2

That's more than the circuit with three capacitors.

So in short, if the capacitors are discharged each time then...

Adding one in series => energy decreased.
Adding one in parallel => energy increased.