Capacitors - Finding the charges and potential at junction

[Saitama]

Homework Statement

Three capacitors of 2μF, 3μF and 5μF are independently charged with batteries of emf's 5V, 20V and 10V respectively. After disconnecting from voltage sources, these capacitors are connected as shown in figure with their positive polarity connected to A and negative polarity earthed. Now a battery of 20V and an uncharged capacitor of 4μF capacitance are connected to junction A as shown with a switch S. When switch is closed, find:

a)the potential of the junction A.

b)final charges on all four capacitors.

(Attachment 1)

Homework Equations

The Attempt at a Solution

Honestly, I have no clue what happens to the initial charges of capacitors when the switch is closed. I tried to solve the problem as if there are no charges on capacitors initially. Attachment 2 shows the simplified circuit I made. I found the potential at A and got 40/7 V which is incorrect, so I think there is some role of the initial charges which I am unable to figure out. :(

Any help is appreciated. Thanks!

 

[SammyS]

Homework Statement

Three capacitors of 2μF, 3μF and 5μF are independently charged with batteries of emf's 5V, 20V and 10V respectively. After disconnecting from voltage sources, these capacitors are connected as shown in figure with their positive polarity connected to A and negative polarity earthed. Now a battery of 20V and an uncharged capacitor of 4μF capacitance are connected to junction A as shown with a switch S. When switch is closed, find:

a)the potential of the junction A.

b)final charges on all four capacitors.

(Attachment 1)

Homework Equations

The Attempt at a Solution

Honestly, I have no clue what happens to the initial charges of capacitors when the switch is closed. I tried to solve the problem as if there are no charges on capacitors initially. Attachment 2 shows the simplified circuit I made. I found the potential at A and got 40/7 V which is incorrect, so I think there is some role of the initial charges which I am unable to figure out. :(

Any help is appreciated. Thanks!

Let's display those circuits as images.

How much charge is on each capacitor immediately after charging?

Once the three capacitors are connected, they are in parallel with with their positive plates connected by conductors. What is the total charge on their positive plates?

 

[Saitama]

Hi SammyS! Nice to see you. :)

How much charge is on each capacitor immediately after charging?

The charges on 2μF, 3μF and 5μF are 10μC, 60μC and 50μC respectively.

Once the three capacitors are connected, they are in parallel with with their positive plates connected by conductors. What is the total charge on their positive plates?

The total charge is 120μC but I don't see how calculating the charges helps.

 

[ehild]

You can determine the potential before closing the switch from that 120 μC.

What happens when the switch is closed? What happens in a very short time after closing the switch? Does the charge stay on the parallel connected capacitors? ehild

 

[Saitama]

Hi ehild! :)

You can determine the potential before closing the switch from that 120 μC.

How?

What happens when the switch is closed? What happens in a very short time after closing the switch? Does the charge stay on the parallel connected capacitors?

I have no clue about this but I guess the charges won't stay on the plates. They should change due to battery.

 

[ehild]

The three capacitors are connected in parallel, and you know the charge of the resultant capacitor, so you can determine the voltage. But the problem does not ask that.

Yes, the charge on the parallel connected capacitors won't stay there. The circuit is grounded, and there is the battery with fixed emf. The charge can flow through the battery and into or out of the Earth .

ehild

 

[Saitama]

Yes, the charge on the parallel connected capacitors won't stay there. The circuit is grounded, and there is the battery with fixed emf. The charge can flow through the battery and into or out of the Earth .

Does that mean that there is no role of the initial charges and I can simply go about solving the circuit as if there were no initial charges?

I am honestly lost. :(

 

[ehild]

I am not sure, either, and without the grounding, the situation would be different.
What would be the potential of A and the charges on the capacitors if it was an ordinary circuit without initially charged capacitors?

What other possibility can you think of?

ehild

 

[Saitama]

What would be the potential of A and the charges on the capacitors if it was an ordinary circuit without initially charged capacitors?

40/7 V would be the potential of A. The charge on 4μF would be 400/7 μC.

What other possibility can you think of?

I can have one of them uncharged initially and others charged, do you ask me this kind of possibility or did I misinterpret your question?

 

[ehild]

40/7 V would be the potential of A. The charge on 4μF would be 400/7 μC.



I can have one of them uncharged initially and others charged, do you ask me this kind of possibility or did I misinterpret your question?

What would be the situation if the initial charge stayed on the three capacitors?


ehild

 

[Saitama]

What would be the situation if the initial charge stayed on the three capacitors?


ehild

With switch S open, the charges on three capacitors would rearrange because the three capacitors are at a different potential from each other.

 

[ehild]

Of course, the charges of the three connected capacitors rearrange, so the potential difference is equal across all of them. You can treat them as a single capacitor, with charge Q=120μC. Now you connect a battery and a 4 μF capacitor across. What will be the potential of A?

You should note that charge can flow through the battery, but can not accumulate in it. So the sum of charges on the left plates of the capacitors is equal to Q=120 μC, x on the lower capacitor and Q-x on the upper one. The right plates will be negatively charged, as the ground provides charge.
Between the ground and point A, the potential changes by equal amount both across the upper capacitor and along the lower branch of the circuit.

ehild

 

[Saitama]

Of course, the charges of the three connected capacitors rearrange, so the potential difference is equal across all of them. You can treat them as a single capacitor, with charge Q=120μC. Now you connect a battery and a 4 μF capacitor across. What will be the potential of A?

You should note that charge can flow through the battery, but can not accumulate in it. So the sum of charges on the left plates of the capacitors is equal to Q=120 μC, x on the lower capacitor and Q-x on the upper one. The right plates will be negatively charged, as the ground provides charge.
Between the ground and point A, the potential changes by equal amount both across the upper capacitor and along the lower branch of the circuit.

ehild

The potential change in the lower branch is: $20-x/4$.
The potential change in the upper branch is: $(Q-x)/10$.

Equating them doesn't give me the correct answer. :(

If I write the potential change in the lower branch as $20+x/4$ and equate, I do get the right answer.

Is this because we assume x to be positive (though it turns out to be negative)?

Also, I don't get why did we solve it this way. Why is it wrong to simply solve it as if there were no charges initially? What would I do if the circuit was not grounded?

 

[ehild]

The potential change in the lower branch is: $20-x/4$.
The potential change in the upper branch is: $(Q-x)/10$.

Equating them doesn't give me the correct answer. :(

Why do you write 20-x/4? Supposing the left plates of capacitors have the charge and the right plates have the opposite charge, UA=x/4+20, UA=(Q-x)/10.
x happens to be negative - it does not change anything.

Also, I don't get why did we solve it this way. Why is it wrong to simply solve it as if there were no charges initially? What would I do if the circuit was not grounded?

The "usual" solution assumes series connection of the capacitors, with equal charges on them which is not true in this case. The initially given charge can not disappear, it only can be distributed between the connected plates.

ehild

 

[Saitama]

The "usual" solution assumes series connection of the capacitors, with equal charges on them which is not true in this case. The initially given charge can not disappear, it only can be distributed between the connected plates.

ehild

Sorry for asking foolish questions but the circuit is earthed, the charge can flow into the ground, right?

Please look at the attached problem. This is a problem from the exam I gave last year.

When switch $S_1$ is pressed and released, the charge on the upper plate of $C_1$ is $2CV_0$. When switch $S_2$ is pressed and released, the charge redistributes between $C_1$ and $C_2$ leaving each of the capacitor's upper plates with charge $CV_0$.

I remember that during the exam, I selected D as one of the correct answers. And D (with B) is the correct answer but where does the charge go? I mean, on the upper plate of $C_2$, the charge was $CV_0$ before $S_3$ was pressed. When $S_3$ is pressed for long time, then according to the given answers, the battery puts new charges on the plates. What happened to the charges that were present before $S_3$ was pressed?

 

[SammyS]

from Post #7:

Does that mean that there is no role of the initial charges and I can simply go about solving the circuit as if there were no initial charges?

I am honestly lost. :(

I'm going back to your question in post #7. Things get muddled after that.

When the switch is open, the battery has no effect on the charges on or voltage across the three initial capacitors.

What is the effective Capacitance of the three capacitors in parallel ?

What is the voltage that an equivalent capacitor with a charge of 120 μC would have across it?

 

[ehild]

I remember that during the exam, I selected D as one of the correct answers. And D (with B) is the correct answer but where does the charge go? I mean, on the upper plate of $C_2$, the charge was $CV_0$ before $S_3$ was pressed. When $S_3$ is pressed for long time, then according to the given answers, the battery puts new charges on the plates. What happened to the charges that were present before $S_3$ was pressed?

The "charge" is not particle, it is an attribute of a piece of material. Usually, materials are electrically neutral, and become charged if they get excess charge - electrons removed or added. A metal piece consists of positive ions and free electrons. The capacitor plate is charged to some positive q - that means it misses some electrons. If you add -q charge (electrons), the plate becomes neutral, all the ions have enough electrons around them. If it is a semiconductor, we say that the added electrons recombine with the holes.
If you add more electrons, the plate becomes negatively charged.
C2 was connected to a battery of opposite polarity. The positive plate got negative charge, the negative plate got positive charge. The added charge neutralized the previous one, and provided excess charge of opposite polarity.

What happens on one plate of the capacitor, the opposite happens on the other plate.

ehild

 

[Saitama]

The "charge" is not particle, it is an attribute of a piece of material. Usually, materials are electrically neutral, and become charged if they get excess charge - electrons removed or added. A metal piece consists of positive ions and free electrons. The capacitor plate is charged to some positive q - that means it misses some electrons. If you add -q charge (electrons), the plate becomes neutral, all the ions have enough electrons around them. If it is a semiconductor, we say that the added electrons recombine with the holes.
If you add more electrons, the plate becomes negatively charged.
C2 was connected to a battery of opposite polarity. The positive plate got negative charge, the negative plate got positive charge. The added charge neutralized the previous one, and provided excess charge of opposite polarity.

What happens on one plate of the capacitor, the opposite happens on the other plate.

ehild

Thanks ehild for the explanation! :)

So the charge stored or present in the battery is responsible for the new charges on plates, am I right in saying this?

I am still not sure about the no earthing case. Can I solve it by assuming as if there were no intial charges?

from Post #7:

I'm going back to your question in post #7. Things get muddled after that.

When the switch is open, the battery has no effect on the charges on or voltage across the three initial capacitors.

What is the effective Capacitance of the three capacitors in parallel ?

What is the voltage that an equivalent capacitor with a charge of 120 μC would have across it?

Hi SammyS!

The effective capacitance is 10μF and voltage is 12V.

 

[gneill]

I don't know if it will help you at this stage of the game, but note that for analysis purposes an initially charged capacitor can be represented by a model consisting of an uncharged capacitor of the same size in series with a voltage supply with a potential difference equal in value to the initial potential across the capacitor.

You can then use all the usual circuit rules to combine sources, etc., to simplify the circuit and determine any change in charge that will occur on the capacitor.

 

[ehild]

So the charge stored or present in the battery is responsible for the new charges on plates, am I right in saying this?

The battery separates the opposite charges, otherwise the whole battery is neutral. The electromotive force of the battery drives positive charges to the three connected capacitors and the same amount of negative charge to the 4μF one.

I am still not sure about the no earthing case. Can I solve it by assuming as if there were no initial charges?

Sorry, I misunderstood something. I did not read the original problem carefully. You get the same result either with earthing or without it. For the non-earthing case you can also apply Gneill's method to find the final charge of the 4 uF capacitor, assuming zero initial charges, but an extra battery.

What I thought was that charge was given only to one plate of all capacitors at the beginning (I saw such problem just about the same time https://www.physicsforums.com/showthread.php?t=734124). Here is a question for you to think about: What is the voltage you measure across the capacitor plates if you give Q charge to one plate, without earthing the other plate? And what is the voltage if the other plate is earthed?

Reading the original problem, the capacitors were charged by connecting them to voltage sources, so one plate got some Q charge, the other plate got -Q charge. The same as if the other plate would have been connected to earth. ehild

 

[Saitama]

For the non-earthing case you can also apply Gneill's method to find the final charge of the 4 uF capacitor, assuming zero initial charges, but an extra battery.

Oh yes, I do get the same answer. Thanks gneill and ehild!

Here is a question for you to think about: What is the voltage you measure across the capacitor plates if you give Q charge to one plate, without earthing the other plate?

The charge rearranges, if the left plate is initially given charge Q, then the outer and inner surfaces of left plate and outer surface of right plate has charge Q/2 and inner surface of right plate has -Q/2. The potential difference is then Q/(2C).

And what is the voltage if the other plate is earthed?

I am not sure but a charge of -Q/2 flows from the Earth to ground surface and potential does not change.

 

[ehild]

Oh yes, I do get the same answer. Thanks gneill and ehild! The charge rearranges, if the left plate is initially given charge Q, then the outer and inner surfaces of left plate and outer surface of right plate has charge Q/2 and inner surface of right plate has -Q/2. The potential difference is then Q/(2C).

I am not sure but a charge of -Q/2 flows from the Earth to ground surface and potential does not change.

If you connect one plate of the capacitor to Earth and give Q charge to the other plate, -Q charge flows to the earthed plate from the ground, and the whole charge appears in the inner sides. The whole electric field is between the plates and it is Q/(Aε), and the potential difference is dQ/(Aε), that is, U=Q/C.

ehild

 

[Saitama]

If you connect one plate of the capacitor to Earth and give Q charge to the other plate, -Q charge flows to the earthed plate from the ground, and the whole charge appears in the inner sides. The whole electric field is between the plates and it is Q/(Aε), and the potential difference is dQ/(Aε), that is, U=Q/C.

ehild

Thanks a lot ehild! :)

BTW, do you know of a book which deals deeply with the concepts of earthing and capacitors?

 

[ehild]

BTW, do you know of a book which deals deeply with the concepts of earthing and capacitors?

No, I do not know any special book about earthing and capacitors, do you really need one?

What you need when solving such problems is only Electrostatics.
As for earth, it is taken as the zero of the potential. When you connect something to the earth, you ensure that it is at the same potential. If a metal plate is earthed, there is zero potential difference between it and the earth. Is there electric field between the plate and the earth?

Place an other metal plate parallel with the earthed one, and give it some positive charge Q. That charge attracts electrons of the earthed plate to its inner surface, making it negatively charged. How much charge has to be accumulated on the earthed plate to ensure zero electric field between it and the earth?

ehild

 

[Saitama]

Sorry for the late reply.

As for earth, it is taken as the zero of the potential. When you connect something to the earth, you ensure that it is at the same potential. If a metal plate is earthed, there is zero potential difference between it and the earth. Is there electric field between the plate and the earth?

Nope, zero potential difference implies no electric field. :)

Place an other metal plate parallel with the earthed one, and give it some positive charge Q. That charge attracts electrons of the earthed plate to its inner surface, making it negatively charged. How much charge has to be accumulated on the earthed plate to ensure zero electric field between it and the earth?

I am not sure but you stated the answer in your previous post. The charge will be zero on the outer surface of the earthed plate and -Q on the inner surface.

I am still in a dilemma. Let none of the plates be earthed. Then the charge distribution is shown in the attachment. When I Earth the right plate, then why doesn't the charge distribution changes according to the attachment 2? I don't see how shifting all the charges to the inner surface (i.e Q on the inner surface of left plate and -Q on the inner surface of right plate) ensure zero potential difference between Earth and the earthed plate (should I say the earthed plate or outer surface of earthed plate?).

 

[ehild]

I don't see how shifting all the charges to the inner surface (i.e Q on the inner surface of left plate and -Q on the inner surface of right plate) ensure zero potential difference between Earth and the earthed plate (should I say the earthed plate or outer surface of earthed plate?).

There is a very primitive answer: opposite charges attract each other. The positive plate attracts electrons from the Earth to the inner surface of the earthed plate: they go as close as possible to the positive plate.
Also, the charge can redistribute in the positive plate. The Earth is at zero potential, but the potential at infinity is also zero. The system has the lowest energy if the field is confined between the plates.
You can place the earthed plate on the ground: It does not change anything: You can imagine that the other plate of the capacitor is a very big metal piece.

Imagine surface charge density σ1 and σ2 on the upper metal plate (σ1+σ2=σ=Q/A) and σ3 (negative) on the surface of the big metal piece. Determine the surface charges knowing that the electric field is zero inside the metals.

εE1=σ1/2+σ2/2-σ3/2
εE2=-σ1/2+σ2/2-σ3/2=0
εE3=-σ1/2-σ2/2-σ3/2
εE4=-σ1/2-σ2/2+σ3/2=0

σ1+σ2=σ --> εE4=-σ/2+σ3/2=0 -->σ3=σ
εE2=-σ1/2+σ2/2-σ3/2=0 --->σ1=0

εE3=-σ1/2-σ2/2-σ3/2 --->σ.

There are surface charges only on the inner surfaces. The electric field between the plates, inside the capacitor is σ/ε, zero outside, (ignoring edge effects).

ehild

 

[Saitama]

$$\begin{array}
\\
εE_1=σ_1/2+σ_2/2-σ_3/2\\
εE_2=-σ_1/2+σ_2/2-σ_3/2=0\\
εE_3=-σ_1/2-σ_2/2-σ_3/2\\
εE_4=-σ_1/2-σ_2/2+σ_3/2=0\\
\\
σ_1+σ_2=σ --> εE_4=-σ/2+σ_3/2=0 -->σ_3=σ\\
εE_2=-σ_1/2+σ_2/2-σ_3/2=0 --->σ_1=0\\
\\
εE_3=-σ_1/2-σ_2/2-σ_3/2 --->σ.\\
\end{array}
$$

I should have written down those equations, I am sorry to ask foolish questions.

That was a very nice explanation, thanks a lot ehild!

 

[Tanya Sharma]

I am still in a dilemma. Let none of the plates be earthed. Then the charge distribution is shown in the attachment. When I Earth the right plate, then why doesn't the charge distribution changes according to the attachment 2?

Because in that arrangement the electric field within the conducting plates will not be zero .