
A parallel plates capacitor is made of square conducting plates of side a and the separation between plates is d. The capacitor is connected with battery of emf V volt as shown in the figure. There is a dielectric slab of dimension with dielectric constant k. At t = 0, dielectric slab is given velocity towards capacitor as shown in the figure. (Neglect the effect of gravity and electrostatic force acting on the dielectric when dielectric is out side of capacitor. Also ignore any type frictional force acting on the dielectric during its motion) let the x be the length dielectric inside the capacitor at t = t sec.


Important Questions on Capacitance

A parallel plate capacitor with square plates is filled with four dielectrics of dielectric constants arranged as shown in the figure. The effective dielectric constant will be:







A parallel-plate capacitor of area , plate separation and capacitance is filled with four dielectric materials having dielectric constants, , and as shown in the figure below. If a single dielectric material is to be used to have the same capacitance in this capacitor, then its dielectric constant is given by


A simple pendulum of mass ' ', length ' ' and charge suspended in the electric field produced by two conducting parallel plates as shown. The value of deflection of pendulum in equilibrium position will be


If is the free charge on the capacitor plates and is the bound charge on the dielectric slab of dielectric constant placed between the capacitor plates, then bound charge can be expressed as :



For a capacitor, the distance between two plates is , the electric field between them is , now dielectric slab having dielectric constant and thickness is placed between them in contact with one plate. In this condition what is the potential difference between its two plates is:



If these two modified capacitors are charged by the same potential , the ratio of the energy stored in the two, would be refers to capacitor and to capacitor ):



