Find the capacitance of a system of three parallel plates, each of area A `metre^(2)` separated by distances `d_(1) and d_(2)` metre respectively. The space between them is filled with dielectrics of relatives dielectric constants `K_(1) and K_(2)`. The dielectric constant of free space is `in_(0)`.

A system of two parallel plates, each of area A, are separated by distance d_(1) and d_(2) . The space between them is filled with dielectrices of permitivities epsilon_(1) and epsilon)_(2) . The permittivity of free space is epsilon_(0) . The equivalent capacitance of the system is

The capacitance of a capacitor, filled with two dielectrics of same dimensions but of dielectric constants K_1 and K_2 respectively as shown will be -

Two parallel plate capacitors, each of capacitance 40 muF, are connected is series. The space between the plates of one capacitor is filled with a dielectric material of dielectric constant K =4. Find the equivalent capacitacne of the system.

The capacitance of a parallel plate capacitor with plate area A and separation d is C . The space between the plates in filled with two wedges of dielectric constants K_(1) and K_(2) respectively. Find the capacitance of resulting capacitor.

A parallel plate capacitor with air between the plates has a capacitance C. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant 6, then the capacitance will become.

A parallel plate capacitor with air between the plates has capacitance of 9pF . The separation between its plates is 'd'. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant k_1=3 and thickness d/3 while the other one has dielectric constant k_2=6 and thickness (2d)/(3) . Capacitance of the capacitor is now