i.) A parallel plate capacitor consists of two circular conducting plates, each

Question

i.) A parallel plate capacitor consists of two circular conducting plates, each with a diameter of 8.69 mm, separated by 1.87 mm of dielectric. When charged to 5.87 x 10-12C the potential difference across the capacitor is 18.2 V. What is the dielectric constant of the material between the plates?

ii.) Four capacitors are connected in parallel. Two of the capacitors are each 20.6 nF. The third capacitor is 14.3 nF. The combination has an equivalent capacitance of 82.2 nF. What is the value of the fourth capacitor?

iii.) Four capacitors are connected in parallel. Two of the capacitors are each 20.6 nF. The third capacitor is 14.3 nF. The fourth capacitor has a capacitance of 73.4 nF. What is the value of the equivalent capacitance?

iv.) A parallel plate capacitor consists of two circular conducting plates, each with an area of 4.73 x 10-6 m2, separated by 1.87 mm of dielectric that has a dielectric constant of 1.21. The potential difference across the capacitor is 18.2 V. How much energy is stored in the capacitor?

Explanation / Answer

a) Q = CV

C = Q/V = 5.87 x 10^-12 / 18.2= 0.322 x 10^-12 C = K ( 8.854 × 10^-12) ( 3.14 x 1.88 x 10^ -5)/ 1.87 x 10^-3

0.322 x 10^ -15 = K ( 10^-17 ) ( 52.26 )

K = 0.61 apprx

b) C4=  82.2 - ( 55.5) = 26.7 nF

c) C eq= 2 (20.6) + 14.3 + 73.4 = 128.9 nF

d) C = 1.21 x 8.854 × 10^-12 x (4.73 x 10-6)/ 1.87 x 10^-3 = 27. 1 x 10^ -15 F apprx

U = 0.5 cv^ 2= 0.5 (  27. 1 x 10^ -15 ) ( 18.2^2) = 4.488 x 10^ -12 J apprx

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