# Finding Capacitance of system

1. May 15, 2015

### Raghav Gupta

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

Four metallic plates each of surface area S are placed at some distance from each other as shown in figure.
Then capacitance of the system between P and Q is

A. $\frac{ε_0S}{3d}$

B. $\frac{ε_0S}{2d}$

C. $\frac{3ε_0S}{2d}$

D. $\frac{3ε_0S}{d}$
2. Relevant equations
Between 2 parallel plate capacitors without dielectric capacitance is given as
$$C = \frac{ε_0A}{d}$$ where A is surface area of plates and d is distance between them.

3. The attempt at a solution
I know if two plates are in series then equivalent capacitance is $\frac{1}{C_{eq}}= \frac{1}{C_1} + \frac{1}{C_2}$
If they are in parallel then equivalent C is C1 + C2.
But I don"t know what the system is in diagram.

Post Script: How do I disable the red lines coming on words below? I am using first time computer for posting a problem in PF.

2. May 15, 2015

### ehild

The middle plates can be doubled and connected, it is the same set-up. So it is three capacitors, like in the figure, all having the same plate surface area. How are the ones with 2d distance connected?

3. May 15, 2015

### Raghav Gupta

How did you convert my diagram to that? Is there some method?

4. May 15, 2015

### ehild

I t
I told you in the previous post. The middle plates can be thought as two plates, connected together by a wire. The arrngement on the left is equivalent with the one on the right: two capacitors, with plates AB and B'C.
Originally, the plates A and B are connected. Do the same on the right ones.

5. May 15, 2015

### Raghav Gupta

Okay, the diagram was the important part.
Now, 2d plates are in parallel
So capacitance is adding both 2 capacitances ε0S/d
Then they are in series with d plate
So capacitance is 1/2 of ε0S/d which is ε0S/2d . So option B is correct.
Thanks ehild.

6. May 15, 2015

### ehild

You are welcome