Series circuit problem - am I correct?

[bcjochim07]

Homework Statement

A simple series circuit consists of a 150 ohm resistor, a 25 V battery, a switch and 2.5 pF parallel plate capacitor(uncharged initially) w/ plates 5mm apart. The switch is closed at t=0.

What is the maximum electric flux and maximum displacement current through the capacitor?

Homework Equations

The Attempt at a Solution

Maximum flux will be achieved when the capacitor is fully charged, at
Vc = Vbat Vc = 25 V Q = (2.5*10^-12)(25) = 6.25 * 10 ^-11 C
Electric flux = EA cos(theta)= qin/epsilon
6.25*10e-11 / 8.85e-12 = <b> 7.06 Vm^2 </b>

Displacement current = 8.85* 10e-12 * dFluxElectric/dt

which = I/epsilon

I need to find the Imax, so I used the formula for the charging of a capacitor:

Q = Qmax(1-e^-t/RC)
I = -dQ/dt = -Qmax/RC * (1-e^-t/RC) RC = 3.75 e-10

the maximum I is Qmax/RC Qmax is calculated above to be 6.25e-11
so Imax is .167 A

so I think the maximum displacement current is .167 A

Did I do this right?

 

[phinds]

so I think the maximum displacement current is .167 A

Got that part right. I didn't check the flux

 

[Delta2]

This looks right to me, just to mention one thing (that it is mentioned in the OP, just not with so big clarity), that the displacement current through the capacitor equals the conventional current that reaches the capacitor's plates. This is a consequence of Gauss's law in integral form by taking the derivative with respect to time in both sides,$$\oint E \cdot dA=\frac{q}{\epsilon_0}\Rightarrow \epsilon_0 \frac{d\oint E\cdot dA}{dt}=\frac{dq}{dt}\Rightarrow \epsilon_0 \frac{d\Phi_E}{dt}=\frac{dq}{dt}$$

 

[Steve4Physics]

Maximum displacement current is $\frac V R = \frac {25V}{150 \Omega} = 0.167A$. It’s as simple as that!

That’s because the displacement current equals the conduction current, as already noted by @Delta in Post #3.

At the instant the voltage is applied, the (uncharged) capacitor offers no opposition to charge-flow: the capacitor acts as a short-circuit (zero resistance). The instantaneous initial circuit current is therefore a maximum and equals $\frac V R$.