The capacitance of a parallel plate air capacitor is 9pF. The separation between the plates is d. The intermidate space is filled up by two dielectric media. The widths of them are `(d)/(3) " and "(2d)/(3)`, and their dielectric constants are 3 and 6, respectively. Find the capacitance of the parallel plate capacitor?

Explanation

As the capacitors are connected in series the equivalent capacitance,
`C_(eq)= (C_1C_2)/(C_1+C_2)" ""……………"(1)`
We know that, `C_1 = k_1(in_0 A)/(d/3)" and "C_2 = k_2(in_0A)/((2d)/(3))`

Putting the values of `C_1 " and C_2` in equation (1),
`C_(eq) = ((in_0A)/((d)/(3))xx(in_0A)/((2d)/(3))xxk_1k_2)/((in_0A)/(d/3)(k_1+(k_2)/(2)))`
`(in_0 A)/(d) = 9pF " and "k_1 =3, k_2 =6`
So, `C_(eq)= (3xx9xx3xx(9)/(2)xx3xx6)/(3xx9(3+(6)/(2)))= 40.5` pF.