Question

4. A dielectric-filled capacitor is directly connected to a 12.0-V battery until it is fully charged. Then the capacitor is disconnected from the battery and, once disconnected, the dielectrie is removed so that air fills the capacitor instead. The dielectric used is pyrex glass, whose dielectric constant is 5.6, while the dielectric constant of air is 1.0059. (a) Determine the voltage across the capacitor plates after the dielectric is removed. (b) With the dielectric removed, the capacitor is discharged by conneeting it to a device with a resistance of 150 Ω Suppose that you would like the capacitor to discharge to 5.0% of its original voltage (the original voltage is the answer to part a) in 3.5 seconds. To achieve this, what should the capacitance be?

Answer 1

4)Given,

V = 12 V ; k = 5.6

We know that,

C = k e0 A/d

V = Q/C

when the battery will be removed the charge will be conseved

V' = Q/C'

after removal C has been decreased by a factor of 5.6

V'/V = Q/C' x C/Q = C'/C = 1.0059/5.6 = 0.179

V' = 0.179625 x 12 = 2.1555 V

Hence, V' = 2.1555 V

b)R = 150 Ohm

5% of 12 = 0.6 V

V = Vc e^-t/RC

0.6 = 12 e^-3.5/RC

0.6/12 = e^-3.5/RC

taking natural log both sides

-2.99 = -3.5/RC

RC = 1.17

C = 1.17/150 = 7.8 milli Farads

Hence, C = 7.8 mF