Each of the three plates shown in figure has an area of 200 cm^2 on one side and the gap between the adjacent plates is 0.2 mm. The emf of the battery is 20V. Find the distribution of charge on various surfaces of the plates. What is the equivalent capacitance of the system between the terminal points?

Explanation

As the potentials of a and c are equal, the capacitors `C_(ab)` and `C_(bc)` are in parallel. Therefore,
`C_(ab)=C_(bc)=(epsilon_(0)A)/d =(epsilon_(0)xx2xx10^(-2))/(0.2xx10^(-3)) = 100 epsilon_(0)(F)`
Equivalent capacitance is
`C=C_(ab)+C_(bc)=(2epsilon_(0)A)/d`
`=(2xxepsilonxx2xx10^(-2))/(0.2xx10^(-3))=200epsilon_(0)(C)`
The charge on plat b is negetive on both faces. Thus, the charge on faces a and c is .
`q_(a)=q_(c)=(epsilon_(0)A)/(d)epsilon`
`=100 epsilon_(0)xx20`
`=2000epsilon_(0)(C)`
`:. Q=2xx2000epsilon_(0)(C)`
`=4000epsilon_(0)(C)`
Alternative method : Let us assume that the potential of the middle plate be zero, Then the potential of the left and right plates will be `20 V` each. Let the positive terminal of the battery lupply a charge `Q` . This charge is divided into two plates as shown in. The charge on the outermost surfaces will be zero,
.
Equal and opposite charges will appear on both surfaces of the mid plate (pate2).The sum of the charges on both the surfaces should be `-Q`, which will flow toward the negative terminal of the battery. we can observe two capacitors `C_(1)` and `C_(2)` for capacitor `C_(1)`
`x=C(20-0)` (i)
For capacitor `C_(2)`
`(Q-x)=C(20-0)` (ii)
From (i)
`x=(epsilon_(0)A)/(d)xx20=(epsilon_(0)(2xx10^(-2)))/(0.2xx10^(-3))xx20=2000epsilon_(0)(C)`
Also from (ii)
`(Q-x)=2000epsilon_(0)(C)`
The net charge on the mid plate is
`-x+(Q-x)=-2xx2000epsilon_(0)=-4000epsilon_(0)(C)`
Hence, the equivalent capacity is
`C_(eq)=Q/(epsilon)=(4000epsilon_(0))/(20)`
or `C_(eq)=200epsilon_(0)(F)`.