A parallel plate capacitor has circular plates of 15.2 cm radius and 0.26 mm sep
ID: 2053241 • Letter: A
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