Question

Please, Help me!

Answer 1

@ganeshie8

Answer 2

@UnkleRhaukus

Answer 3

how many capacitors do u see?

Answer 4

@mashy 3?

Answer 5

@Luigi0210 ....

Answer 6

I have never seen such an arrangement before, but that doesn't mean it doesn't exist.

I am assuming the charging source is connected between plates P and R. Now the plate Q is placed between the two plates that are going to be charged by source......but is not connected to anything else (no reference to the charging source. So what does a "floating plate do between two plates which have charging source connected. the only thing that I can think of it would cause the distance to be 2d. I can't see it increasing the area as it is not connected to any part of the circuit that is charging the circuit. Sooooo if I were to make a WAG at this I would say the second choice is the best one.

Answer 7

I would also be interested in what mashy has to say.

Answer 8

\[C=\frac{Q}{| \Delta V|}=\frac{\epsilon_0A}{d}\]

Answer 9

I am not familiar with the set-up either, but you can use the idea that potentials add in series
\[|\Delta V|=|\Delta V_1|+|\Delta V_2|; Q_{eq}=\frac{Q}{C_1}+\frac{Q}{C_2}\]
Capacitance in Parallel
\[C = C_1+C_2 ... \]

Answer 10

@astrophysics

Answer 11

correction
\[\frac{Q}{C_{eq}}=\frac{Q}{C_1}+\frac{Q}{C2}..\]