For Problems 5 to 6: Figure shows a capacitor connected to a voltage source. The

Question

For Problems 5 to 6: Figure shows a capacitor connected to a voltage source. There are two dielectric slabs stacked in the capacitor. The dielectric slabs are not perfect dielectrics, thus they have finite conductivities. Hint: Notice that electric flux densities in dielectrics 1 and 2 are equal: D-D2 Another hint: You can imagine this structure as two capacitors connected in series. Can you find the voltage Kon capacitor 1. V-4cos(at),o-10,ad /sec, w-o1m, 1:0.ln?, ,-3,e,2-5, ?,-100,?,-200,di-d,-0.002m 82,02 Problem [51 Calculate the magnitude of the total conduction current in dielectric slab1 Problem [6] Calculate the magnitude of the total displacement current in dielectric slab Problem [7] 10 points> In a medium the electric field intensity is given as E-5sin (6t). Calculate the magnitude of the magnetic flux density at (rs 4,0 ,2-7, t-0)

Explanation / Answer

given

Capacitor, connected to a voltage source

dielectric 1:

d1, epsilon1, sigma1

dielectric 2:

d2, epsilon2, sigma2

V = 4cos(wt)

w = 10^6 rad/s

w' = 0.1 m

l = 0.1 m

er1 = 3

er2 = 5

sigma1 = 100

sigma2 = 200

d1 = d2 = 0.002 m = d

5. these can be considered two resistors in series

hence

net resistance

R = d/w'*l*sigma1 + d/w'*l*sigma2 = (1/sigma1 + 1/sigma2)*d/w'*l

R = 0.003 ohm

hence

total conduction current in dielectric slab 1, i1 = V/R = 1333.333333*cos(wt) A

6. total dislacement current = id

id = (dQ/dt)

now

Capacitance of the total capacitor = C

C = (d/w'*l*epsilon1 + d/w'*l*epsilon2)^-1

C = (w'*l/d)(1/epsilon1 + 1/epsilon2)^-1 = 83.00625*10^-12 F

hence

Q = CV

Q = 83.0062*10^-12*4*cos(wt) = 332.025*10^-12*cos(wt) C

hence

id = 0.000332025 *sin(wt) A

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