(a) Find the potential difference between the plates of a parallel plate capacitor of capacitance `5muF` if a charge `+800muC` is placed on one plate and-`200muC` on another plate.
(b) The charege given to plate are as shown. The capacitance between the adjacent plates is `C`. find the charge on outer surface of plate `3`, and potential difference between plate `1` and `2`.
(c) If `A` is area of each plate and `d` is separation between adjacent plates, find the charge supplied by battery.
(d) Five idential capacitor plates each of area `A` are arranged such that adjacent plates are at a distance `d` apart. the plates are connected to a source of emf `V` as shown. what is the magnitude and nature of charge on plates `1` and `3` respectively?

Explanation

(a) In case of isolated plates having charges, the charge on outer surfaces of extreme plates will be
`(Q_(1)+Q_(2)+…+Q_(n))/(2)`
as explained earlier.

change on outer surface of plates, `(Q_(1)+Q_(2))/(2)`
Charge distribution:

here, capacitor is having charge : `((Q_(1)-Q_(2))/(2))`
`p.d`. between plates :
`((Q_(1)-Q_(2))/(2))//C=(Q_(1)-Q_(2))/(2C)`
here, `Q_(1)=800muC,Q_(2)=-200muC`, `C=5muF`
p.d. `V=(Q_(1)-Q_(2))/(2C)=(800-(-200))/(2xx5)=100V`
(b)
The charge on outer surfaces of extreme plates will be
`(Q+(-Q)+3Q)/(2)=(3Q)/(2)`
Charge distribution:

Charge on outer surface of plate `3` is `(3Q)/(2)`.
`p.d`. between plates `1` and `2`:
`V=(Q//2)/(c)=(Q)/(2C)`
(c)
First, rearrange the plates.

`C_(eq)=(3C)/(2)=(3Aepsilon_(0))/(2d)`
(d)
First rearrange the plates.

Charge on plate `1:-CV=(-Aepsilon_(0)V)/(d)`
Charge on plate `3:CV+CV=2 CV=(2Aepsilon_(0)V)/(d)`