# 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...