- The plates of a parallel plate capacitor are separated by 1cm .Two parallel sided dielectric slabs of thickness 0.6cm and 0.4cm fill the space between the plates.The dielectric constants of the two slabs are 3 and 5 respectively and a potential difference of 210V is applied across the plates.Electric field in the slab of 0.6cm thick is `(k)/(-)times10^(4)`V/m .The value of 'k` is

The plates of a parallel plate capacitor are separated by 1cm .Two parallel sided dielectric slabs of thickness 0.6cm and 0.4 cm fill the space between the plates.The dielectric constants of the two slabs are 3 and 5 respectively and a potential difference of 210V is applied across the plates.Electric field in the slab of 0.6cm thick is (k)/(2)times10^(4)V/m .The value of k is

The plates of a parallel plate capacitor are separated by a distance d = 1 cm . Two parallel sided dielectric slabs of thickness 0. 7cm and 0.3cm fill the space between the plates. If the dielectric constants of the two slabs are 3 and 5 respectively and a potential difference of 440V is applied across the plates. Find (a) The electric field intensities in each of the slabs. (b) The ratio of electric energies stored in the first to that in the second dielectric slab.

The capacitance of a capacitor between 4//3 times its original value if a dielectric slab of thickness t=d//2 is inserted between the plates (d is the separation between the plates). What is the dielectric consant of the slab?

The separation between the plates of a parallel plate capacitor is d and the area of each plate is A. iff a dielectric slab of thickness x and dielectric constant K is introduced between the plates, then the capacitance will be

Capacitance of a capacitor becomes 7/6 times of its original value if a dielectric slab of thickess t=2/4d is introduced in between the plates, d is the separation between the plates. The dielectric constant of the dielectric slab is

The plates in a parallel plates capacitor are seperated by a distance d with air as the medium between the plates. In order to increase the capacity by 66% a dielectric slab of thickness (3d)/5 is introduced between the plates. What is the dielectric constant of the dielectric slab ?

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 is connected across a source of covstant potential difference. If a dielectric slab is introduced between the two plates, then

The plates of parallel plate capacitor are charged upto 100V . A 2mm thick dielectric slab is inserted between the plates.Then to maintain the same potential difference,the distance between the plates is increased by 1.6mm . The dielectric constant of the slab is.