Student Workbook for Physics for Scientists and Engineers: A Strategic Approach, Vol 1. (Chs 1-21)
4th Edition
ISBN: 9780134110646
Author: Randall D. Knight (Professor Emeritus)
Publisher: PEARSON
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Chapter 23, Problem 9CQ
A sphere of radius
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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 rh has charge -Q. The electric field E at a radial distancer 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 = (-aaoe-r/ao + B
+ bo
By definition, the capacitance C is related to the charge and potential difference by:
C= Q
I Vab
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 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/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 = (-aager/ao + B
+ bo
By definition, the capacitance Cis 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 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 (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:
['´e
Vob =
Edr= -
Edr
Calculating the antiderivative or indefinite integral,
Vab = (-aaoe¯r7ao + 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 / (
(erb/ao - eralao) + B In(
) + bo (
))
Chapter 23 Solutions
Student Workbook for Physics for Scientists and Engineers: A Strategic Approach, Vol 1. (Chs 1-21)
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