Questions about a capacitor with 4 parallel plates

[palaphys]

Homework Statement

attached below

Relevant Equations

C=Q/V $1/C_{net} = 1/C_1 + 1/C_2$

Till now, all I have found out, is that for option P, it matches with $C_0 /3$.
not able to understand, how to find the capacitance, if two plates are "shorted".

My attempt (to solve Q): I mentally connected $S_2 to S_3$ with a connecting wire, to ensure that the two plates are at the same potential.
would that make two capacitors?? Like $S_1 + S_2$as a single capacitor, which is (probably) in series with $S_3 +S_4$, which yields, $C_{net} = C_0/2$

however,
1. I am unsure whether this logic is correct
2. How to apply the same, to solve R and S? not able to solve them.

 

[.Scott]

Your reasoning is "so far, so good".
If you look at your choices (A, B, C, and D) what you're left to solved is only S.

For S, how many capacitors do you have in effect? How are they arranged? Draw a schematic for yourself. It might be useful to think of each plate as two parallel connected plates facing away from each other. For each pair of half-plates facing across a gap to each other, you have another capacitor.

 

[kuruman]

When $S_2$ is shorted with $S_3$, the surfaces are at the same potential. This is the same situation as if a conducting plate of thickness $d$ were inserted halfway between $S_4$ and $S_1$ as shown in the attached schematic. Then you can remove part of the conducting material and match the conventional depiction of series capacitors.

(Edited to add schematic.)

 

[palaphys]

Your reasoning is "so far, so good".
If you look at your choices (A, B, C, and D) what you're left to solved is only S.

For S, how many capacitors do you have in effect? How are they arranged? Draw a schematic for yourself. It might be useful to think of each plate as two parallel connected plates facing away from each other. For each pair of half-plates facing across a gap to each other, you have another capacitor.

When $S_2$ is shorted with $S_3$, the surfaces are at the same potential. This is the same situation as if a conducting plate of thickness $d$ were inserted halfway between $S_4$ and $S_1$ as shown in the attached schematic. Then you can remove part of the conducting material and match the conventional depiction of series capacitors.
View attachment 360265
(Edited to add schematic.)

how to apply this logic to solve options R and Q? not sure about those

 

[.Scott]

how to apply this logic to solve options R and Q? not sure about those

For Q, you have @kuruman schematic - with, effectively (as you have described), 2 capacitors in series.
You probably meant R and S , there are exactly 3 capacitors: S1/S2, S2/S3, and S3/S4. Separate those out in a schematic and then connect them up. You will see series and parallel circuits. Then apply your series and parallel equations.

Then reply here with as far as you were able to go.
The key is for us to direct you through the process - not to do the work ourselves.

 

[kuruman]

how to apply this logic to solve options R and Q? not sure about those

The logic is this.

When you are asked about the capacitance between two plates you first imagine that the two plates are connected to a battery. This establishes a potential that varies linearly in the region between plates (electric field is uniform and non-zero) and is constant (electric field is zero) in the region outside the plates. The equipotentials are planes parallel to the plates.

Now consider how the potential between the plates connected to the battery will change if you introduce additional conducting surfaces as indicated by the question. Hint: Assume that one plate is at potential $+V$, the other plate is at potential $-V$ and the plane equidistant to the plates is at zero potential.

 

[palaphys]

Then reply here with as far as you were able to go.
The key is for us to direct you through the process - not to do the work ourselves.

ok. here is my attempt for S

also I think it is valid to "break" S2 and S3 and connect them with a connecting wire, since they would be at the same potential and hence nothing would change (stored charge would be the same)
now, i marked the potentials as x and 0, wherever applicable(with positive terminal of battery connected to S1, and -ve terminal to s2, as the option S asks for the capacitance between S1 and S2.). not sure how to proceed, but hope I am right till now

 

[palaphys]

ok. here is my attempt for S
View attachment 360310
also I think it is valid to "break" S2 and S3 and connect them with a connecting wire, since they would be at the same potential and hence nothing would change (stored charge would be the same)
now, i marked the potentials as x and 0, wherever applicable(with positive terminal of battery connected to S1, and -ve terminal to s2, as the option S asks for the capacitance between S1 and S2.). not sure how to proceed, but hope I am right till now

OH wait, are these in parallel? it seems so. guess the answer is $3C_0$

 

[.Scott]

Very good !! You did it.

 

[palaphys]

thanks for the help. @kuruman @.Scott

 

[kuruman]

ok. here is my attempt for S

How about your attempt for R?

 

[palaphys]

How about your attempt for R?

solved it, two came out to be parallel with the other one, so I got $2C_0 / 3$

 

[kuruman]

solved it, two came out to be parallel with the other one, so I got $2C_0 / 3$

Think again. $S_1$ nd $S_3$ are connected to a battery. $S_2$ and $S_4$ are connected to each other. What is in parallel?

 

[palaphys]

Think again. $S_1$ nd $S_3$ are connected to a battery. $S_2$ and $S_4$ are connected to each other. What is in parallel

I meant something like this

 

[kuruman]

Please redraw your idea showing the battery connections with a "+" and a "-" and also the shorted plates. Besides, you are showing six plates when there are only four.