# Homework Help: A problem about capacitor

1. Nov 19, 2015

### gracy

1. The problem statement, all variables and given/known data
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. Relevant equations

$V$=$\frac{Q}{C}$

3. The attempt at a solution

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.

Last edited: Nov 19, 2015
2. Nov 19, 2015

### ehild

What have you learnt 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?

3. Nov 19, 2015

### Staff: Mentor

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

4. Nov 19, 2015

### gracy

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

5. Nov 19, 2015

### andrevdh

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

6. Nov 19, 2015

### Staff: Mentor

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...

7. Nov 19, 2015

### gracy

Two capacitors or two plates of a capacitor?

8. Nov 19, 2015

### BvU

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 ?

9. Nov 19, 2015

### gracy

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

10. Nov 19, 2015

### 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.

11. Nov 19, 2015

### gracy

Am I going in right/correct direction?

12. Nov 19, 2015

### Staff: Mentor

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

13. Nov 19, 2015

### gracy

$E$=$\frac{σ}{2ε0}$
Right?

14. Nov 19, 2015

### Staff: Mentor

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?

15. Nov 20, 2015

### gracy

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

$E_2$=$\frac{σ2}{2ε0}$
Vector addition of $E_1$ and $E_2$

16. Nov 20, 2015

### Staff: Mentor

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

17. Nov 20, 2015

### gracy

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

18. Nov 20, 2015

### Staff: Mentor

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?

19. Nov 20, 2015

### gracy

$\frac{Q1}{2ε0s}$-$\frac{Q2}{2ε0s}$

20. Nov 20, 2015

### Staff: Mentor

Keep going...

21. Nov 20, 2015

### gracy

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

22. Nov 20, 2015

### Staff: Mentor

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.

23. Nov 20, 2015

### gracy

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

24. Nov 20, 2015

### Staff: Mentor

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

25. Nov 20, 2015

### gracy

Electric field multiplied by r?