The plates of parallel plate capacitor A consist of two metal discs of identical
ID: 1333274 • Letter: T
Question
The plates of parallel plate capacitor A consist of two metal discs of identical radius, R1 = 2.67 cm, separated by a distance d = 1.49 mm, as shown in the figure.
a) Calculate the capacitance of this parallel plate capacitor with the space between the plates filled with air.
b) A dielectric in the shape of a thick-walled cylinder of outer radius R1 = 2.67 cm, inner radius R2 = 1.17 cm, thickness d = 1.49 mm, and dielectric constant = 3.17 is placed between the plates, coaxial with the plates, as shown in the figure. Calculate the capacitance of capacitor B, with this dielectric.
c) The dielectric cylinder is removed, and instead a solid disc of radius R1 made of the same dielectric is placed between the plates to form capacitor C, as shown in the figure. What is the new capacitance?
Explanation / Answer
a)
Area of each disk, A = pi*R1^2
= pi*(2.67*10^-2)^2
= 2.24*10^-3 m^2
C = A*epsilon/d
= 2.24*10^-3*8.854*10^-12/(1.49*10^-3)
= 1.331*10^-11 F <<<<<<-----------Answer
b)
here It acts as two capacitirs connected ia prallel.
Cnet = C1(dielectric) + C2(air)
C1(dielctric) = K*A*epsilon/d
= k*pi*(R1^2 - R2^2)*epsilon/d
= 3.17*pi*(2.67^2 - 1.17^2)*10^-4*8.854*10^-12/(1.49*10^-3)
= 3.4*10^-11 F
C2(air) = A*epsilon/d
= pi*R2^2*epsilon/d
= pi*1.17^2*10^-4*8.854*10^-12/(1.49*10^-3)
= 2.56*10^-12 F
Cnet = C1 + C2
= 3.4*10^-11 + 0.256*10^-11
= 3.656*10^-11 F <<<<<<<<<<----------------Answer
c) Need fugure to solve this part.
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