A parallel plate capacitor is to be designed with a voltage rating 1kV using a material of dielectric constant 10 and dielectric strength `10^(6)V/m` .What minimum area of the plates is required to have a capacitance of `88.5pF` ?

A parallel plate capacitor is to be designed with a voltage rating 1 KV using a material of dielectrical constant 3 and dielectric strength about 10^(7) Vm^(-1) . [Dielectric strength is the maximum electric field a material can tolerate without break down, i.e, without starting to conduct electrically through partial ionisation. For safety, we should like the field never to exceed say 10% of the dielectric strength]. What minimum area of the plates is required to have a capacitance of 50 pF ?

A parallel plate capacitor is charged to a certain voltage. Now if the dielectric material (with dielectric constant k ) is removed then the

A parallel plate capacitor is to be prepared by using plates of the same area, of one of the dielectric given below: Which dielectric gives the maximum capacitance?

A parallel plate capacitor is filled with dielectrics as shown in Fig. What is its capacitance ?

A parallel plate capacitor having a separation between the plates d, plate area A and material with dielectric constant K has capacitance C_(0) . Now one-third of the material is replaced by another material with dielectric constant 2K, so that effectively there are two capacitors one with area 1/3A, dielectric constant 2K and another with area 2/3A and dielectric constant K. If the capacitance of this new capacitor is C then (C)/(C_(0) is