A parallel plate capacitor has circular plates of 15.2 cm radius and 0.26 mm sep

Question

A parallel plate capacitor has circular plates of 15.2 cm radius and 0.26 mm separation.

a) Calculate the capacitance. Put answer in the form of "a.bc x 10^(y)" pF?.

b) What charge will appear on the capacitor plates if a potential difference of 13.8 V is applied to it? Put answer in the form of "a.bc x 10^(y)" nanocoulomb.

please help solve step by step
c) If the region between the plates is now filled with material having a dieletric constant of 4.2, what is the new value of the capacitance? put answer in the form of "a.bc x 10^(x)"pF

Explanation / Answer

a) C = A/d

A = r^2 = .0725 m^2

d = .26*10^-3

C = 8.84 * 10^-12 F

b)Charge on capacitor plates = CV = 8.84 * 10^-12 * 13.8 = 1.2201 * 10^-10 C

c) Capacitance with dielectric

4.2 * C = 4.2 * 8.84 * 10^-12 = 3.7128 * 10^-11 F

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