What is the capacitance of the system between P and Q?

[Raghav Gupta]

Homework Statement

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. $\frac{3ε_0S}{d}$

Homework 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.

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.

 

[ehild]

Homework Statement

View attachment 83496

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. $\frac{3ε_0S}{d}$

Homework 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.

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.

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?

 

[Raghav Gupta]

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?

View attachment 83500

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

 

[ehild]

I t

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

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.

 

[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.

 

[ehild]

You are welcome