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 = 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 R1made 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, A = pi*R^2
= pi*(0.0267 m)^2
=2.24*10^-3 m^2

d = 1.49mm = 1.49*10^-3 m

C= ebsoleneo*A/d
=(8.854*10^-12)*(2.24*10^-3) /(1.49*10^-3)
=1.33*10^-11 F

b)
There will 2 capacitor,1 with dielectric and other without it connected in parallel.

with out dielectric:
Area, A = pi*R2^2
= pi*(0.017 m)^2
=9.08*10^-4 m^2

d = 1.49mm = 1.49*10^-3 m

C1= ebsoleneo*A/d
=(8.854*10^-12)*(9.08*10^-4) /(1.49*10^-3)
=5.4*10^-12 F

With dielectric:
Area, A = pi*R1^2 - pi*R2^2
= pi*(0.0267)^2 - 9.08*10^-4 m^2
=1.33*10^-3 m^2

d = 1.49mm = 1.49*10^-3 m

C2= k*ebsoleneo*A/d
=3.17*(8.854*10^-12)*(1.33*10^-3) /(1.49*10^-3)
=2.5*10^-11 F

These 2 are present as if connected in parallel
Cnet =C1+C2 = 5.4*10^-12 + 2.5*10^-11 =3*10^-11 F

c)
Area, A = pi*R1^2
= pi*(0.0267)^2
=2.24*10^-3 m^2

d = 1.49mm = 1.49*10^-3 m

C= k*ebsoleneo*A/d
=3.17*(8.854*10^-12)*(2.24*10^-3) /(1.49*10^-3)
=4.22*10^-11 F

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.