Q57 of 94 Page 166

A capacitor is formed by two square metal-plates of edge a, separated by a distance d. Dielectrics of dielectric constants K1 and K2 are filled in the gap as shown in figure. Find the capacitance.


These two capacitors are connected in series.



To find out the capacitance, let us consider a small capacitor of


differential width dx at a distance x from


the left end of the capacitor.


The two capacitive elements of dielectric


constants K1 and K2 are with plate


separations as -


and in series,


respectively as seen from fig.


Also, differential plate areas of the capacitors are adx.


We know, capacitance c is given by-



Where,


A= Plate Area


d= separation between the plates,


∈0 = Permittivity of free space = 8.854 × 10-12 m-3 kg-1 s4 A2


k = dielectric strengthof the material


Then, looking into the fig, the capacitances of the capacitive elements of the elemental capacitors are given by –


and



We know that equivalent capacitance of capacitors connected in


series is given by the expression –





Now, integrating both sides to get the actual capacitance,







Looking back into the fig.



Substituting in the expression for capacitance C,



Now,





More from this chapter

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55

A parallel-plate capacitor having plate area 400 cm2 and separation between the plates 1.0 mm is connected to a power supply of 100V. A dielectric slab of thickness 1.0 mm and dielectric constant 5.0 is inserted into the gap.

a) Find the increase in electrostatic energy.


b) If the power supply is now disconnected and the dielectric slab is taken out, find the further increase in energy.


c) Why does the energy increase in inserting the slab as well as in taking it out?


 56

Find the capacitances of the capacitors shown in figure. The plate area is A and the separation between the plates is d. Different dielectric slabs in a particular part of the figure are of the same thickness and the entire gap between the plates is filled with the dielectric slabs.


 58

Shows two identical parallel plate capacitors connected to a battery through a switch S. Initially, the switch is closed so that the capacitors are completely charged. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant 3. Find the ratio of the initial total energy stored in the capacitors to the final total energy stored.


 59

A parallel-plate capacitor of plate area A and plate separation d is charged to a potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. Find the work done on the system in the process of inserting the slab.