A parallel-plate capacitor has plate area 20 cm2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 kΩ resistor. Find the energy for the capacitor 8.9 μs after the connections are made.

Energy stored in a capacitor:


The energy stored in a capacitor with capacitance C , charge is given by:



where V is the potential difference across the capacitor.


Capacitance of a Capacitor in presence of a dielectric:
The capacitance of the capacitor is initially C0 and then a dielectric medium of dielectric constant K is inserted between the plates. The new capacitance is



Also for parallel plate capacitors,



Where ϵ0 is the permittivity of free space, A is the area of plate and l is the distance between the plates.


Charging a capacitor:


A capacitor of capacitance C is being charged using a battery of emf ϵ through a resistance R . A switch S is also connected in series with the capacitor. The switch is initially open. The capacitor is uncharged at first. At t=0, the switch is closed. The current through the circuit at anytime t>0 is given by:



Where I0 is the initial current.


The charge is given by:



Note that is known as time constant.


Given,


Area of the plate,


(


Now, distance of separation:


Dielectric constant,


Emf of battery,


Resistance ,


Time,


Now,





Time constant,


We want to find the charge on the capacitor at t = 8.9μs





Now, energy of the capacitor is given by:





Hence, the capacitor stores of energy after 8.9μs.


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