Q. 534.0( 2 Votes )

A parallel-plate capacitor of capacitance 5 μF is connected to a battery of emf 6V. The separation between the plates is 2 mm.

a) Find the charge on the positive plate.

b) Find the electric field between the plates.

c) A dielectric slab of thickness 1 mm and dielectric constant 5 is inserted into the gap to occupy the lower half of it. Find the capacitance of the new combination.

d) How much charge has flown through the battery after the slab is inserted?

Answer :

Given-


Capacitance C=5 μF = F


Voltage, V=6v


Separation between plates, d=2 mm=2×10-3 m


a)The charge on the positive plate is calculated using


q = c×v


where, c is the capacitance


and v = voltage applied


Thus,


q=5 μF×6 V


=30 μC


b) The electric field between the plates of the capacitor is given by



Where, v is the applied voltage and d is the distance between the capacitor plates



=3×103 V/m


c)Given-


Distance between the plates of the capacitor, d =2×10-3 m
Dielectric constant of the dielectric material inserted, k = 5
Thickness of the dielectric material inserted, t = 1×10-3 m


capacitance of the capacitor= 5 μF


Now,
To calculate area of the plates of the capacitor,



Where,


A = area


k = dielectric constant of the material placed


d= separation between the plates


substituting the values,





When the dielectric placed in it, the capacitance becomes



t=thickness of the material


substituting the values,




1)


d)
The charge stored in the capacitor initially is -


C=5×10-6 F


Also, V=6 V


Now, we know



where v is the applied voltage and c is the capacitance




Now, when the dielectric slab is inserted ,charge on the capacitor, from 1)



Charge, Q’





Now, charge flow is given by,



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