A pallel -plate capacitor whith the plate area `100cm^2` and the separation between the plate 1.0cm is connected across a battery of emf 24 volts .Find the force of attraction between the plates.

A parallel plate air capacitor has capacity 'C' distance of separation between plates is 'd' and potential difference 'V' is applied between the plates force of attraction between the plates of the parallel plate air capacitor is :

What is the area of the plates of a 2 F parallel plate capacitor given that the separation between the plates is 0.5 cm ?

In a parallel plate capacitor with air between the plates, each plate has an area of 6 xx 10^(-3)m^(2) and the distance between the plates is 3 mm. Calculate the capacitance of the capacitor. If this capacitor is connected to a 100 V supply, what is the charge on each plate of the capacitor?

A battery is used to charge a parallel plate capacitor till the potential difference between the plates become equal to the electromotive force of the battery. The ratio of the energy stored in the capacitor and the work clone by the battery will be .......

What is the area of the plates of a 2 F parallel plate capacitor, given that the separation between the plates is 0.5 cm? [You will realise from your answer why ordinary capacitors are in the range of µF or less. However, electrolytic capacitors do have a much larger capacitance (0.1 F) because of very minute separation between the conductors.]

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 charge will be :

A slab of material of dielectric constant K has the same area as the plates of a parallel plate capacitor but has a thickless (1/2)d . Where d is the separation of the plates . How is the capacitance changed when the slab is inserted between the plates

A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness (3/4)d, where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

A slab of material of dielectric constant K has the same area as the plates of a parallel-plate capacitor but has a thickness ((3)/(4)) d where d is the separation of the plates. How Is the capacitance changed when the slab ls inserted between the plates ?