Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rp has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = ae-r/ao + B/r + bo where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vah between the two conductors. The potential difference is related to the electric field by: Vab = Edr Calculating the antiderivative or indefinite integral, Vab = (-aage-r/ao0 + B + bo By definition, the capacitance C is related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (e-fb/a0 - efalao) + B In( ) + bo ( ))

Physics for Scientists and Engineers: Foundations and Connections
1st Edition
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Katz, Debora M.
Chapter28: Current And Resistance
Section: Chapter Questions
Problem 69PQ
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Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius rą has a charge of +Q, while the outer cylinder of radius rh has charge -Q. The electric field E at a radial
distance r from the central axis is given by the function:
E = ae-rlao + B/r + bo
where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance
First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by:
Vah
Edr =
Edr
Calculating the antiderivative or indefinite integral,
Vab = (-aaoe-r/ao + B
+ bo
By definition, the capacitance C is related to the charge and potential difference by:
C =
Evaluating with the upper and lower limits of integration for Vab, then simplifying:
C = Q/(
(e-rb/ao - e-Talao) + ß In(
) + bo (
))
Transcribed Image Text:Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius rą has a charge of +Q, while the outer cylinder of radius rh has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = ae-rlao + B/r + bo where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Vah Edr = Edr Calculating the antiderivative or indefinite integral, Vab = (-aaoe-r/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (e-rb/ao - e-Talao) + ß In( ) + bo ( ))
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