Shows a parallel plate capacitor with circular plates of radius r_0 & area A = p

Question

Shows a parallel plate capacitor with circular plates of radius r_0 & area A = pi(r_0) = 100 cm^2 There is air between the plates. The capacitance C = 30 pF (30 Times 10^-12F). It is being charged by a V0 = 70 V battery through a resistor of R = 2.0 Ohm (NOTSHOWN in the future1). At the instant the battery is connected (time t = 0), the initial charge on the plates is zero. It is at that same instant when the electric field between the plates is changing most rapidly. As time goes on, the charge Q on the plates increases with time towards a maximum value Q_max,. Assume that the electric field E between the plates is spatially uniform and that it is zero anywhere outside the edges of the plates. Calculate: The maximum charge Q_max which will be on the plates after a very long time has passed. The time constant tau for this RC circuit. The current I flowing into the plates at t = 0. For the following, recall that the formula for the time dependent charge Q(t) on the capacitor plates in an RC circuit is Q(t) = CV_0[1 - exp(-t/tau)J. Also recall that the electric field between the plates of a parallel plate capacitor is E = (sigma/epsilon_0), where sigma is the surface charge density (Q/A) on the plates, Write a formula for the time rate of change (dE/dt) of the electric field between the plates. Calculate the numerical value of this quantity at t = 0. Write a formula for the Maxwell Displacement Current between the plates. Calculate the numerical value of this displacement current at t = 0. Assuming that the field lines associated with the magnetic field B induced between the plates are circles perpendicular to the electric field E, as in Fie. 2. write a formula for the time dependence of this magnetic field at the outside edge (at r = r_0) of the plates Calculate the numerical value of this field at t = 0.

Explanation / Answer

We know that the charge is given by

Q = CV

Q = 30 *70 = 0.21 nC

b) The time constant is given by

T = RC

T = 2 *30 = 60 ps (pico second)

c) Intially the capacitor acts as a conducting wire hence the initla current is given by

I = V/R = 70/2 = 35 A

d) We know that

E = V/d

And V = Q/C , here in this circuit Q varies as

Hence

dE/dt = dQ/det *(1/Cd)

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

dQ/dt = Qo/RC * (e^( t/RC)

Hence

dE/dt = Qo/RdC^2 * (e^( t/RC)

(Please post other parts seperately)

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