Derive expresion for capacitance of a parallel plate capacitor and explain the combination of capacitors in series.
Or.
Write expression for three capacitors in series and parallel combinations.

Explanation

Capacitors in Series. Capacitors are said to be connected in series when charge proceeds from one point to the other through a single path.
If figure `C_(1),C_(2), . . ,C_(n)` etc, are connected in series. Let cahrge +q be given to the left plate of `C_(1)`, a charge-q is induced on inner side of right plate of `C_(1) and +q` on the outer side of this plate. charge thus flows from left to rightl. let `V_(1),V_(2),. . .V_(n)` be the potential differences between two sides of `C_(1),C_(2), . . ,C_(n)` respectively V be the total P.D. across all the plates.

`therefore V_(1)=(q)/(C_(1)),V_(2)=(q)/(C_(2)), . . .,V_(n)=(q)/(C_(n)) and V=(q)/(C_(s))`
where `C_(s)` is the resultant capacitance of the arrangement in series.
Now `V=V_(1)+V_(2)+ . .+V_(n)`
or `(q)/(C_(s))=(q)/(C_(1))+(q)/(C_(2))+ . .+(q)/(C_(n))`
or `(1)/(C_(s))=(1)/(C_(1))+(1)/(C_(2))+ . .. +(1)/(C_(n))`
NOTE: if we have two capacitors of capacitance `C_(1) and C_(2)`, then `(1)/(C_(s))=(1)/(C_(1))+(1)/(C_(2))`
Capacitors in parallel. capacitors are said to be connected in parallel if the .positively charged. plates of all the capacitors are connected together at a point A while the .earthed. plates are connected together at another point B. the point B is also connected to earth while the source of charge is connected between the points A and B (fig).

Capacitors having capacities `C_(1),C_(2), . .. ,C_(n)`, draw charges `q_(1),q_(2), . . . ,q_(n)` in accordance with their capacities. if q is the total charge drawn from the source, then
`q=q_(1)+q_(2)+ . . .+q_(n)` . . (1)
Since all the capacitors are connected between two common points A and B therefore, the potential difference across each of them is the same i.e. V. this is also the potential difference across the two terminals of the source of charge.
When `q_(1)=C_(1)V`,
`q_(2)=C_(2)V`,
. . . .
. . . ..
. . . . . . .
`q_(n)=C_(n)V`.
If `.C_(p).` is the capacity of the combination, then
`V=(q)/(C)`
or `q=CV`
Putting the values of `q,q_(1),q_(2), . ..,q_(n)` in Eq. (i),
we get
`CV=C_(1)V+C_(2)V+ . . .+C_(n)V`
or `C_(p)=C_(1)+C_(2)+ . .. . +C_(n)`
Note. For three capacitors, we shall use only three capacitors `C_(1),C_(2) and C_(3)`, then
`(1)/(C_(s))=(1)/(C_(1))+(1)/(C_(2))+(1)/(C_(3))`
and `C_(p)=C_(1)+C_(2)+C_(3)`.