If a dielectric slab of thickness 5 mm and dielectric constant K=6 is introduced between the plates of a parallel plate air capacitor, with plate separation of 8 mm, then its capacitance is

A dielectric of thickness 5cm and dielectric constant 10 is introduced between the plates of a parallel plate capacitor having plate area 500 sq. cm and separation between the plates 10cm. The capacitance of the capacitor with dielectric slab is (epsi_(0) = 8.8 xx 10^(-12)C^(2)//N-m^(2))

A dielectric slab is placed between the plates of a parallel plate capacitor. Its capacitance.

A dielectric slab is placed between the plates of a parallel plate capacitor. Its capacitance

A parallel plate air capacitor of plate separation d is charged to a potential difference triangle V . A slab of thickness d and dielectric constant k is introduced between the plates while the battery remains connected. Find the ratio of energy stored in the capacitor after and before the introduction of the dielectric slab.

A parallel plate capacitor with a dielectric slab with dielectric constant k= 3 filling the space between the plates is charged to potential V and isolated. Then the dielectric slab is drawn out and another dielectric slab of equal thickness but dielectric constant k'= 2 is introduced between the plates. The ratio of the energy stored in the capacitor later to that initially is:

Separation between the plates of parallel plate capacitor is 5 mm. this capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3mm and dielectric constant K=10 is placed between the plates as shown. potential difference between the plates after the dielectric slab has been introduced is-

Separation between the plates of parallel plate capacitor is 5 mm. this capacitor, having air as the dielectric medium between the plates, is charged to a potential difference 25 V using a battery. The battery is then disconnected and a dielectric slab of thickness 3mm and dielectric constant K=10 is placed between the plates as shown. potential difference between the plates after the dielectric slab has been introduced is-