Capacitor Formation with Unequal Charges: A Potential Difference Dilemma

[gracy]

Homework Statement

Two plates (area=s) charged to +q1 and +q2 (q2<q1)are brought closer to form a capacitor of capacitance C.The potential difference across the plates is

2. Homework Equations
$V$=$\frac{Q}{C}$

The Attempt at a Solution

[/B]
I don't know how are these plates going to form a capacitor as both of them have same type of (i.e positive)charge and in order to form a capacitor two conducting plates with equal and opposite charges are required.

 

[ehild]

What have you learned about a parallel-plate capacitor? What is the capacitance if the area of the plates is s and the distance between them is d? Does the capacitance depend on the charge of the plates?

 

[gneill]

I would be inclined to look at the net electric field between the plates.

 

[gracy]

I would be inclined to look at the net electric field between the plates.

$\vec{E}$=$\frac{σ}{ε0}$=$\frac{Q}{ε0A}$

 

[andrevdh]

What if one would to consider two capacitors one charged with a q1 charge and the other with a q2 charge?

 

[gneill]

$\vec{E}$=$\frac{σ}{ε0}$=$\frac{Q}{ε0A}$

That's the idea, but check your formula for the electric field... a sheet of charge has two sides so you need to take that into account when you apply Gauss' Law to find the field.

edit: See for example: Hyperphysics: Electric Feild, Flat Sheets of Charge

Now, you have two sheets of charge with different charge densities...

 

[gracy]

two capacitors

Two capacitors or two plates of a capacitor?

 

[BvU]

Homework Statement

Two plates (area=s) charged to +q1 and +q2 (q2<q1)are brought closer to form a capacitor of capacitance C
...
in order to form a capacitor two capacitors conducting plates with equal and opposite charges are required.

Does the capacitance depend on the charge of the plates

Let q1 = 1, q2 = 2 for simplicity.

If the plates are far apart (far enough that they don't influence each other significantly), can you imagine the charges are evenly distributed over the surfaces ?

What happens if they are brought closer together ?

 

[gracy]

What happens if they are brought closer together ?

Capacitance increases.
I don't want to sound skeptical that's why I avoided question mark!It does not mean I am confident about it.

 

[gracy]

My thought process is
For a parallel plate capacitor
$C$=$\frac{Aε0}{d}$
As d (distance between the two plates )decreases,C i.e Capacitance should increase.

 

[gracy]

Am I going in right/correct direction?

 

[gneill]

Am I going in right/correct direction?

I still feel that you'd be better to look at the net field between the plates. Gauss' Law will give you the answer.

 

[gracy]

check your formula for the electric field..

$E$=$\frac{σ}{2ε0}$
Right?

 

[gneill]

$E$=$\frac{σ}{2ε0}$
Right?

That will give you the field between the plates due to one plate with net charge density σ. You have two plates with different σ's. Draw a sketch and pay attention to the field directions. How do they sum?

 

[gracy]

Electric field due to one plate having charge density σ1 in between the two plates(we will avoid electric field at other places)
$E_1$=$\frac{σ1}{2ε0}$

$E_2$=$\frac{σ2}{2ε0}$

How do they sum?

Vector addition of $E_1$ and $E_2$

 

[gneill]

Okay, so assuming that you choose an appropriate direction for the net field, what's an expression for its magnitude?

 

[gracy]

what's an expression for its magnitude?

$\frac{σ1}{2ε0}$-$\frac{σ2}{2ε0}$
as Q1>Q2
Hence σ1>σ2

 

[gneill]

Good. But you can expand that a bit using the problem's given variables: s, q1, q2. Usually you want to express any results using the given parameters. You can pretty it up by factoring out the common variables.

Next, given a constant electric field, how do you find the potential difference between two locations?

 

[gracy]

$\frac{Q1}{2ε0s}$-$\frac{Q2}{2ε0s}$

 

[gneill]

Keep going...

 

[gracy]

Next, given a constant electric field, how do you find the potential difference between two locations?

But it is not mentioned that electric field is constant,is it?

 

[gneill]

But it is not mentioned that electric field is constant,is it?

The field from a relatively large sheet of charge is uniform. Here "relatively large" means that for the location of interest the distance from a face of the sheet is small compared to the size of the sheet, and it's not too close to the edges of the sheet where "edge effects" interfere with the uniformity.

 

[gracy]

So I should consider electric field to be uniform in this case?

 

[gneill]

So I should consider electric field to be uniform in this case?

Yes, clearly. There are two parallel sheets of charge on plates forming a capacitor.

 

[gracy]

how do you find the potential difference between two locations?

Electric field multiplied by r?

 

[gracy]

Answer given in my book is
$\frac{q1-q2}{2C}$

 

[gneill]

Electric field multiplied by r?

Sure. You make it look like a guess though. It should be an expression that's given in your text and notes (although they're more likely to use "d" rather than "r").

Answer given in my book is
$\frac{q1-q2}{2C}$

That's fine. Once you've factored and grouped the variables for the potential difference expression you should be able to pick out the expression for capacitance in the mix and make the substitution.

 

[gracy]

Ok.
$V$=$\frac{q1-q2}{2sε0}$d

But we know $C$=$\frac{sε0}{d}$

Here s=A=Area

$\frac{d}{sε0}$=$\frac{1}{C}$

$V$=$\frac{q1-q2}{2C}$

Thanks @gneill.

 

[gneill]

You're welcome!