What is the Charge on a Capacitor with Given Area and Plate Separation?

[gracy]

Homework Statement

find the charge on each plate given area is a and separation between two consecutive plate is d

Homework Equations

$Q$=$C$$V$

The Attempt at a Solution

I don't know how to proceed from here.

 

[gneill]

You need another relevant equation. What is the capacitance of a parallel plate capacitor with plate area $a$ and plate separation $d$?

 

[gracy]

What is the capacitance of a parallel plate capacitor with plate area aa and plate separation dd?

$C$=$\frac{aε0}{d}$

 

[gneill]

Yes. Proceed.

 

[gracy]

Q=C$\frac{aε0}{d}$
Will this be charge on each plate?

 

[gneill]

Q=C$\frac{aε0}{d}$
Will this be charge on each plate?

No. That has two capacitance terms. You need a voltage and a capacitance: Q = CV.

 

[gracy]

Haste makes waste!
Q=V$\frac{aε0}{d}$
Will this be charge on each plate?

 

[gneill]

Haste makes waste!
Q=C$\frac{aε0}{d}$
Will this be charge on each plate?

Still too hasty. You've written the same thing again

Each capacitor plate in your circuit diagram will have the same magnitude of charge on it. You'll have to sort out the signs of the charges. Then determine how to locate those charges on the plates of the original figure -- remember, one physical plate is shared between two capacitors.

 

[gracy]

Each capacitor plate in your circuit diagram will have the same magnitude of charge on it

But hint says plate B would have different charge.

 

[gneill]

But hint says plate B would have different charge.

Yes. I presume "plate B" is the middle plate of the physical setup? You didn't label your diagram.

Note that I said: "Each capacitor plate in your circuit diagram will have the same magnitude of charge on it". There is a distinction between the circuit diagram and the physical arrangement in that the circuit diagram shows two plates in two separate capacitors whereas the physical setup uses a shared plate. Re-read the entire contents of my post #8.

 

[gracy]

You didn't label your diagram.

 

[gracy]

Each capacitor plate in your circuit diagram will have the same magnitude of charge on it. You'll have to sort out the signs of the charges. Then determine how to locate those charges on the plates of the original figure -- remember, one physical plate is shared between two capacitors.

Should not plate B cntain zero charge as it is earthed?

 

[gneill]

View attachment 92767

Should not plate B cntain zero charge as it is earthed?

No, "earth" is a vast pool of available charges, both positive and negative and overall neutral. It can supply whatever charge is required to respond to electrostatic forces. Or, if you like, it can source or sink any amount of electrons, if you want to go with an atomic model for the plates and conductors.

Your charge placement is not correct. Note that plates A and C are symmetrical and connected to the same potential (V). They should end up with similar charges distributed in the same way. Charges are made available to those plates via their connection to V. So far I think we've been assuming that V is a positive potential with respect to ground (earth), which is fine, but you should state that assumption in your solution.

 

[gracy]

that V is a positive potential with respect to ground

I did not understand that.

 

[gneill]

I did not understand that.

V is taken to be a voltage supply. Voltage is a potential difference. When a terminal is just labeled "V" then it is assumed that that potential difference is between the terminal and the ground reference.

 

[gracy]

Note that plates A and C are symmetrical and connected to the same potential (V). They should end up with similar charges distributed in the same way

I don't quite get the relationship between being symmetrical and connected to the same potential and having same charge distribution.
I distributed charge by only considering charge induction.

 

[cnh1995]

But hint says plate B would have different charge.

Plates 2 and 3 along with the wire joining them, form plate B in your original question.

Q=Vaε0d\frac{aε0}{d}

That is for one capacitor(1-2 or 3-4). The capacitors are in parallel. So you can find the total charge now.

 

[gneill]

I don't quite get the relationship between being symmetrical and connected to the same potential and having same charge distribution.

If you rotate the drawing on the page, do you expect the charges on the plates to change? Is there any distinction (electrically) between the two outer plates?

I distributed charge by only considering charge induction.

You'll have to explain how you ended up with different distributions on the two outer plates.

 

[cnh1995]

I don't quite get the relationship between being symmetrical and connected to the same potential and having same charge distribution.
I distributed charge by only considering charge induction.

A and C are connected to same potential V and you have shown net +ve charge on A and other plates neutral. Why A? Why not C?

 

[gracy]

Is it correct now?

 

[cnh1995]

View attachment 92767

Should not plate B cntain zero charge as it is earthed?

Its 'grounded' i.e it is at 0 potential. It doesn't have to be an actual earthing. It can be battery's -ve terminal.

 

[cnh1995]

Is it correct now?
View attachment 92773

Now you've shown all the plates neutral. There should be net charge on each plate.

 

[gracy]

V is taken to be a voltage supply. Voltage is a potential difference. When a terminal is just labeled "V" then it is assumed that that potential difference is between the terminal and the ground reference.

As in here V is potential difference between A and B

Potential difference is always between two points
But in the following image(my op)there is only one lead connected and it is written "V" so in such cases it is assumed that other point is earth."V "is potential difference between the one terminal shown and the earth.Right?

 

[gracy]

There should be net charge on each plate.

Just because question says find the" charge".If this question appeared as MCQ and option was to be zero then?

 

[gracy]

It doesn't have to be an actual earthing. It can be battery's -ve terminal.

I did not understand.

 

[gneill]

Is it correct now?
View attachment 92773

No. Charges will be "pushed onto" the outer plates by the voltage supply, and "pushed off" the middle plate via the ground connection. The plates will not remain neutral since charge can enter or leave via their external connections. In other words, there's no reason to balance the charges on the faces of any given plate so that they will sum to zero.

And again, the charge distributions on the outer plates should be symmetric with respect to the middle plate.

 

[cnh1995]

Just because question says find the" charge".If this question appeared as MCQ and option was to be zero then?

You yourself correctly interpreted the circuit as two capacitors in parallel connected across some voltage V. So the plates MUST have a net charge.

 

[gneill]

But in the following image(my op)there is only one lead connected and it is written "V" so in such cases it is assumed that other point is earth."V "is potential difference between the one terminal shown and the earth.Right?

Right. Here's an alternative depiction of your circuit:

 

[cnh1995]

I did not understand.

The symbol used is normally called 'ground' i.e. a reference point in the circuit. Normally, it is the negative terminal of the battery (exceptions do exist), however, you can choose any point in the circuit as reference. All the voltages are measured w.r.t. that point. Earthing is done from protection point of view. Anyways, this is not the point of the discussion, so I won't say much.

 

[gracy]

I still don't understand why plates can not be neutral.