Home
Class 11
PHYSICS
The plates in a parallel plates capacito...

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 ?

Promotional Banner

Similar Questions

Explore conceptually related problems

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

Explain why the capacitance of a parallel plate capacitor increases when a dielectric slab is introduced between the plates.

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?

There is an airfilled capacitor of capacity C. when the plate separation is doubled and a dielectric is introduced between the plates, the capacitance becomes 2 C. the dielectric constant of the dielectric is

A parallel plate capacitor has a capacity C. The separation between the plates is doubled and a dielectric medium is introduced between the plates. If the capacity now becomes 2C, the dielectric constant of the medium is

The plates of a parallel plate capacitor are charged to 100V. Then a 4mm thick dielectric slab is inserted between the plates and then to obtain the original potential difference , the distance between the system plates is increased by 2.00 mm. the dielectric constant of the slab 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 becomes 7/6 times its original value if a dielectric slab of thickness t=(2)/(3) is introduced between its plates, where d is the separation between its plates, what is the dielectric constant of the slab?

A parallel plate capacitor carries a harge Q. If a dielectric slab with dielectric constant K=2 is dipped between the plates, then