A parallel plate capacitor of value 1.77 mu F is to be made using a dielectric material (dielectric constant 2000 ,breakdown strength of `3times10^(6)Vm^(-1)`).In order to make such a capacitor, which can withstand a potential difference of 40V across the plates of the separation

The dielectric to be used in a parallel-plate capacitor has a dielectric constant of 3.60 and a dielectric strength of 1.60 xx 10^7 V//m . The capacitor is to have a capacitance of 1.25 xx10^-9F and must be able to withstand a maximum potential difference of 5500 V . What is the minimum area the plates of the capacitor may have?

A parallel plate capacitor is to be designed with a voltage rating 1 KV using a material of dielectric 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 ionization. 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 ?

The parallel combination of two air filled parallel plate capacitor of capacitance C and nC is connected to a battery of voltage, V. When the capacitors are fulley charged, the battery is removed and after that a dielectric material of dielectric constant K is placed between the two plates of the first capacitors. the new potential difference of the combined system is

A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K= (5)/(3) is inserted between the plates, the magnitude of the induced charge will be :

Two identical parallel plate capacitors are connected in series to a battery of 100V . A dielectric slab of dielectric constant 4.0 is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively

Two identical parallel plate capacitors are connected in series and then joined in series with a battery of 100 V . A slab of dielectric constant K=3 is inserted between the plates of the first capacitor. Then, the potential difference across the capacitor will be, respectively.

A parallel plate capacitor of capacity 5 mu F and plate separation 6 cm is connected to a 1V battery and is charged. A dielectric of dielectric constant 4 and thickness 4 cm is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is.

A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by K=K_(0)(1+alphax). . Calculate capacitance of system: (given alpha d ltlt 1) .

A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by K=K_(0)(1+alphax). . Calculate capacitance of system: (given alpha d ltlt 1) .

A parallel plate capacitor with plate area A & plate separation d is filled with a dielectric material of dielectric constant given by K=K_(0)(1+alphax). . Calculate capacitance of system: (given alpha d ltlt 1) .