A parallel-plate capacitor of plate area 40 cm2 and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V through a 16 Ω resistor. Find the electric field in the capacitor 10 ns after the connections are made.
Formulas/Concepts Used:
Capacitance:
If two conductors have a potential difference V between the them and have charges Q and -Q respectively on them, then their capacitance is defined as
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:
Emf of battery,
Resistance ,
Time,
Now,
Time constant,
We want to find the charge on the capacitor at t = 10ns
Now,