Capacitor Plate Charges After Switch is Closed

[Saitama]

Homework Statement

Plates of a capacitor have charge 2CE and CE initially. Now the switch S is closed. Which of following statements is true? Assume that the battery, resistor and wires do not have any capacitance.

A)There is no change in charge at outer surfaces of the capacitor.
B)There is no change in charge at the inner surfaces of the capacitor.
C)Charge flowing through the battery is zero.
D)Charge flowing through the battery is CE/2.

Homework Equations

The Attempt at a Solution

When the switch S is open, the charges rearrange as shown in attachment 2.

The set up is at a potential E/2 so some charge must flow to bring the capacitor at potential E when the switch is closed. The magnitude of final charges on the inner surface is CE. Hence a charge of CE/2 flows through the battery.

But I am unsure what happens to the charge on outer surfaces. I have no idea about what to do with them. :(

Any help is appreciated. Thanks!

 

[TSny]

For arbitrary battery voltage, what general results can you deduce regarding the charges on the outer surfaces of the two plates? Remember that the electric field inside the conducting material of the plates must be zero everywhere.

Also, what happens to the total charge of both plates added together when the switch is closed?

 

[Saitama]

For arbitrary battery voltage, what general results can you deduce regarding the charges on the outer surfaces of the two plates? Remember that the electric field inside the conducting material of the plates must be zero everywhere.

Let an uncharged capacitor of capacitance C be connected to a battery of emf E. The charge appearing on the inner surfaces is CE and -CE. I don't know if it is possible to determine the magnitude of charges on the outer surfaces but I have deduced that charges on the outer surface must be of equal magnitude and same sign.

Also, what happens to the total charge of both plates added together when the switch is closed?

The total charge changes but honestly I still don't see what happens to charges on outer surfaces.

 

[TSny]

The total charge changes but honestly I still don't see what happens to charges on outer surfaces.

Think of the battery as just moving charge from one plate to the other plate. So, how does the total charge of the system (both plates together) change when the switch is closed?

 

[Saitama]

Think of the battery as just moving charge from one plate to the other plate. So, how does the total charge of the system (both plates together) change when the switch is closed?

Now I am even confused by the attempt. Initially, the charges are shown in attachment 2. When a battery is connected, charge CE and -CE appears on the inner surfaces. I said that a charge CE/2 flows through battery but that is only possible when the charges on the outer surfaces do not change. How do I show that charges don't change on the outer surfaces?

 

[TSny]

Use what you said you have shown:

(1) Charges on the two outer surfaces must be of equal magnitude and same sign.
(2) Charges on the two inner surfaces must be of equal magnitude and opposite sign.

And did you come to a definite conclusion about whether or not the total charge (on all surfaces added together) changes when the switch is closed?

 

[Saitama]

And did you come to a definite conclusion about whether or not the total charge (on all surfaces added together) changes when the switch is closed?

Initial charges (from left to right): 3CE/2,CE/2,-CE/2,3CE/2.

If I assume that the charge on outer surfaces do not change, then final charges are: 3CE/2, CE,-CE,3CE/2.

The total charge stays the same but how to show that charge on outer surfaces do not change? I am sorry for silly questions. :(

 

[TSny]

Since the charges on the inner surfaces are equal and opposite, they add to zero. What must be the total charge of the two outer surfaces?

 

[Saitama]

What must be the total charge of the two outer surfaces?

For the given case, it is 3CE.

 

[TSny]

Right. Since the total charge on the two inner plates is always zero, the total charge of the system is the same as the total charge on the outer plates. But, you know that the total charge of the system doesn't change when closing the switch. So, what can you say about the total charge on the outer plates when the switch is closed?

 

[Saitama]

But, you know that the total charge of the system doesn't change when closing the switch.

Sorry, my knowledge has become a little rough on electromagnetism as its quite some time I touched this topic.

Why the total charge of the system doesn't change, the battery can provide additional charges. I was thinking that since the outer surface is directly connected to the wires, charges could flow and the charge on outer surface might change. Why the charges don't flow out in the circuit?

 

[TSny]

The battery moves charge from one plate to the other, but it does not store charge. So, any charge that leaves one plate and moves to the battery must be accompanied by an equal amount of charge that moves from the battery to the other plate.

 

[Saitama]

The battery moves charge from one plate to the other, but it does not store charge. So, any charge that leaves one plate and moves to the battery must be accompanied by an equal amount of charge that moves from the battery to the other plate.

But how does this explains that the amount of charge stays the same on outer surfaces?

 

[mfb]

But how does this explains that the amount of charge stays the same on outer surfaces?

The total charge of the setup is constant, you cannot have a total charge on the inner surfaces (both added together), and you have an additional condition for the charge distribution on the outer surfaces (see the previous posts).

 

[Saitama]

The total charge of the setup is constant, you cannot have a total charge on the inner surfaces (both added together), and you have an additional condition for the charge distribution on the outer surfaces (see the previous posts).

Sorry this is going to be a dumb question.

Initially total charge of the system is 3CE+Q, where Q is the charge stored in battery. When the switch is closed, a charge CE/2 flows to the capacitor. The charge remaining in battery is Q-CE/2. And charge on inner surfaces of capacitor is CE and -CE. The net charge on system is now 3CE+Q-CE/2 and this is not equal to the charge initially present in system.

 

[mfb]

There is no total charge stored in the battery. It always receives as much charge on one side as it gives away on the other, so the battery stays neutral.

When the switch is closed, a charge CE/2 flows to the capacitor.

Why?

 

[Saitama]

Why?

To bring the capacitor to a potential equal to that of battery, right?

Can you please have a look at this thread too: https://www.physicsforums.com/showthread.php?t=719805 ?

 

[mfb]

To bring the capacitor to a potential equal to that of battery, right?

The battery has no capacitance, so it has no relevant potential.

 

[Saitama]

The battery has no capacitance, so it has no relevant potential.

Can you please explain more? Can you please give me a link? I don't know what I am missing here.

If we initially had uncharged plates and connected them to battery, there would be no charge on the outer surface? I still cannot see why the charges from the outer surface flow into the circuit when there is a current flowing in the circuit. I know that current flows only for a very small time but still, I can't grasp this. :(

 

[mukundpa]

I think the confusion is due to the terms potential and potential difference. A battery is creating a potential difference between the two plates. As in steady state there is no current (flow of charges) through the circuit, the potential difference between the plates will be equal to the EMF of the battery.

 

[mukundpa]

oh you have gone :-)

Initially the potential difference across the plates is E/2 (that is because of the inner charges). As the battery is connected the charges CE/2 will flow through the battery to make inner charges +CE and -CE and the potential difference E between the plates, equal to the emf of the battery. the charges on outer surfaces remains unchanged.

 

[Saitama]

oh you have gone :-)

Initially the potential difference across the plates is E/2 (that is because of the inner charges). As the battery is connected the charges CE/2 will flow through the battery to make inner charges +CE and -CE and the potential difference E between the plates, equal to the emf of the battery. the charges on outer surfaces remains unchanged.

I guess I will have to keep a note of this.

Anyways, thank you TSny, mfb and mukundpa!