Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outercylinder 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-ao + B/r + bowhere alpha (a), beta (8), 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= -EdrCalculating the antiderivative or indefinite integral,Vab = (-aaoe-r/a0 + B+ boBy 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 - eTalao) + B Inc) + bo ())

Given data The capacitor consists of two coaxial thin cylindrical coordinates. The radius of the…

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 (8), 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: Vab= "Edr °Edr Calculating the antiderivative or indefinite integral, Vab = (-aage-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 Vgab, then simplifying: C= Q/( (e-rb/ao -eTalao) + B In( ) + bo ( ))

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 (8), 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: Vab= "Edr °Edr Calculating the antiderivative or indefinite integral, Vab = (-aage-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 Vgab, then simplifying: C= Q/( (e-rb/ao -eTalao) + B In( ) + bo ( ))

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter20: Electric Potential And Capacitance

Section: Chapter Questions

Problem 22P

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