Consider a parallel-plate capacitor constructed from two circular metal plates o
ID: 2004865 • Letter: C
Question
Consider a parallel-plate capacitor constructed from two circular metal plates of radius R. The plates are separated by a distance of 1.5 mm.(a) What radius must the plates have if the capacitance of this capacitor is to be 2.6 µF?
m
(b) If the separation between the plates is decreased, should the radius of the plates be increased or decreased to maintain a capacitance of 2.6 µF
(c) Find the radius of the plates that gives a capacitance of 2.6 µF for a plate separation of 3.0 mm.
m
(d) If the space between the plates of the capapcitor is now filled with mica, what will be the new capacitance?
µF
Explanation / Answer
Let the radius of circular plates = R The separation between the plates, d = 1.5 mm = 1.5 * 10^-3 m A) The capacitance, C = 2.6 F = 2.6 * 10^-6 F We have a formula for the capacitance, C = A/d C = R^2/d R = sqrt(Cd/) = 11.85 m B) If the separation between the plates is decreased then the radius will increased to maintain the capacitance 2.6 F C) If the separation, d = 3 mm = 0.003 m We have, R = sqrt(Cd/) = 9.45 m D) When the mica is placed between the plates then the formula for the capacitance, C = k * 2.6 F Where k is the dielectric constant of mica and is given by the value 5.4 Then the new capacitance, C = 5.4 * 2.6 * 10^-6 C = 14.04 * 10^-6 F = 14.04 FRelated Questions
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