The plates of parallel plate capacitor A consist of two metal discs of identical

Question

The plates of parallel plate capacitor A consist of two metal discs of identical radius, R1 = 3.67 cm, separated by a distance d = 3.67 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 = 3.67 cm, inner radius R2 = 2.17 cm, thickness d = 3.67 mm, and dielectric constant ? = 3.13 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?

Please write answers clearly

The plates of parallel plate capacitor A consist of two metal discs of identical radius, R1 = 3.67 cm, separated by a distance d = 3.67 mm, as shown in the figure. Calculate the capacitance of this parallel plate capacitor with the space between the plates filled with air. A dielectric in the shape of a thick-walled cylinder of outer radius R1 = 3.67 cm, inner radius R2 = 2.17 cm, thickness d = 3.67 mm, and dielectric constant ? = 3.13 is placed between the plates, coaxial with the plates, as shown in the figure. Calculate the capacitance of capacitor B, with this dielectric. 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) capacitance of air filled capacitor

C0=8.85*10^-12*3.14*(3.67*10^-2)^2/3.67

=102.03*10^-13= 10.203 pF

b) area covered by dielectric
A'=3.14(R1^2-R2^2)
=3.14(R1+R2)(R1-R2)
=3.14(3.67+2.17)(3.67-2.17)*10^-4
=27.5*10^-4
therefore capacitance of a part of capacitor containing dielectric
C'=3.13*8.85*10^-12*A'/d
=207.5*10^-13=20.7 pF
capacitance of remainng part of the capacitor
C''=8.85*10^-12[3.14*(3.67*10^-2)^2-A']/d  
=0.356 pF
since both the parts are in parallel combination therefore
capacitance C=C'+C''=20.7+0.356= 21.056

c) in part c the space between the plates is completely filled with dielectric
therefore capacitance
C=dielectric constant*capacitance of air filled capacitor
=3.13*10.203=32 pF

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