A battery 60 V initially uncharged capacitor 20 mu F resisto

A battery (6.0 V), initially uncharged capacitor (20 mu F), resistor (150 Ohm) and a switch are connected in series. The switch is closed at t = 0. How much charge is on the capacitor at time t = 4.1 ms? After a long time (t = t_L), a dielectric (with constant k) is suddenly inserted between the plates. Which of the listed quantities will be different*: just before vs. just after the dielectric is inserted? (Select all that apply.) charge on each plate voltage difference across the plates net electric field between the plates electric field due to the plates alone potential energy stored on the capacitor just before the dielectric is inserted, vs. a long time after (t = t_LL) the dielectric remains inserted? (Select all that apply.) charge on each plate voltage difference across the plates net electric field between the plates electric field due to the plates alone potential energy stored on the capacitor * if it changes, you should be able to figure out if it increases or decreases, and by what factor (see HW#6, Prob #9!) Not for credit: Sketch a graph of all of these quantities vs. time, from t = 0 to t_LL. Also make graphs vs. time of the voltage drop across, current through, and power dissipated by the resistor.

Solution

a)

The charge on the capacitor is given by:

Q = Qo*(1-e^(-t/RC))

where Qo = C*V

= 20*10^-6*6

= 1.2*10^-4 C

So, at t = 4.1 ms

Q = 1.2*10^-4*( 1 - e^(-0.0041/(150*20*10^-6)))

= 8.94*10^-5 C

= 89.4 uC <-----answer

b)

Due to introduction of dielectric, E(net Electric field) also changes <----ans

c)

Charge(Q) remains constant before and after the dielectric is introduced

Now, Q = C*V

As the dielectric is introduced, C increases, thus V must decrease to make Q constant

So, voltage difference across plates changes <-----ans

V = Ed

So, E = V/d

As V changes, E(net Electric field) also changes <----ans

net electric potential energy stored ,

U = Q^2/2C

As U is proportional to C, so, U also changes <------- ans