4. A parallel plate capacitor is constructed with a 50 um thick ceramic dielectr

Question

4. A parallel plate capacitor is constructed with a 50 um thick ceramic dielectric with relative permittivity of 15 and dielectric breakdown of 25 MV/m. a. What is the area of the capacitor if the measured capacitance is 100 pF? b. What is the maximum voltage that can be applied? c. How much charge is stored on the capacitor at the maximum rated voltage? If you need the capacitor to withstand twice as much voltage, but have the same capacitance value, what variables need to change? You cannot use a different ceramic dielectric material. d. Protective Coating Dielectric - Ceramic Disc Electrode Connecting Wire TE

Explanation / Answer

Given

parallel plate capacitor with

dielectric material of thickness is d= 50*10^-6 m

permitivity value is k = 15

dielectric breakdown is V = 25 MV/m

a.

Area of the capacitor is A with Capaictance C = 100 pF

we know that C = epsilon0*A/d

100 *10^-12 = 8.854*10^-12*A/(50*10^-6)

A = 0.00056472 m^2

b. Maximum voltage can be applied is V = E*d

V = 25*10^6*50*10^-6 = 1250 V

c.

Charge Q = C*V

Q = 100*10^-12*1250 C = 1.25*10^-7 C = 0.125*10^-6 C

d.

to have double the withstand voltage decreasing the thickness of the dielectric we can achieve

Maximum voltage can be applied is V = E*d

2V = E*d

2V = 25*10^6*d

2500 = 25*10^6*d

d = 0.0001 m = 100*10^-6 m = 2d

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