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 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 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 = (-aaoer/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C= Q Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (erb/a0 - eralao) + 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 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 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 = (-aaoer/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C= Q Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q/( (erb/a0 - eralao) + B In( ) + bo ( ))
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Transcribed Image Text:Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius rg 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-r/a0 + 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:
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 = Q
Evaluating with the upper and lower limits of integration for Vab, then simplifying:
C = Q / (
(erb/a0 - eTa/a0) + B In(
)+ bo (
))
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