We now have two quite different expressions for the line integral of the magnetic field around the same loop. the point here is to see that they both are intimately related to the charge q(t) on the left capacitor plate. first find the displacement current idisplacement(t) in terms of q(t).

q(t) =CV(t) = έAV(0)/d sin ωt

Step-by-step explanation:

Step 1: External electromagnetic force is equal to the voltage of the capacitor . The electric field between the capacitor plates =

E(t) = έ(t)/d = V(0) sin (ωt)/d

Step 2: we calculate the electrical flux between the capacitor= φ(t)=AE(t) = AV(0) sin wt/d

Step 3 : The displacement current can be gotten from the equation below;

I displacement(t) = E(0) dφ(t)/DT= E(0) AV(0)w/d cos wt

Step 4: The current at the outside terminals of the capacitor is the addition of the current used to charge the capacitor and the current in the resistor.

The charge on the capacitor is;

q(t) =CV(t) = έAV(0)/d sin ωt...

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the correct answer is $7.38

step-by-step explanation: