Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
9th Edition
ISBN: 9781305266292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 25, Problem 68AP
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(b)
To determine
The proof for
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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 rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function:
E = αe-r/a0 + β/r + b0
where alpha (α), beta (β), a0 and b0 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:
First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by:
Calculating the antiderivative or indefinite integral ,
Vab = (-αa0e-r/a0 + β + b0 )
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/a0 - e-ra/a0) + β ln() + b0 () )
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 rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function:
E = αe-r/a0 + β/r + b0
where alpha (α), beta (β), a0 and b0 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:
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 rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function:
E = αe-r/a0 + β/r + b0
where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance.
Chapter 25 Solutions
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
Ch. 25.1 - two points and are located within a region in...Ch. 25.2 - QUICK QUIZ 24.2 The labeled points in Figure 24.4...Ch. 25.3 - In Figure 24.8b, take q2, to be a negative source...Ch. 25.4 - In a certain region of space, the electric...Ch. 25 - Prob. 1OQCh. 25 - Prob. 2OQCh. 25 - Prob. 3OQCh. 25 - Prob. 4OQCh. 25 - Prob. 5OQCh. 25 - Prob. 6OQ
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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 rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 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: Calculating the antiderivative or indefinite integral , Vab = (-αa0e-r/a0 + β + b0 ) 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/a0 - e-ra/a0) + β ln() + b0 () )arrow_forwardHow many electrons are in a 1.00-g electrically neutral steel paper clip? The molar mass of steel is approximately that of iron, or 55.845 g/mol, and a neutral iron atom has 26 electrons.arrow_forwardE = E1 E = 0 E= E1 E= E1 The electric field is measured all over a cubical surface, and the pattern of field detected is shown in the figure above. On the right side of the cube, the electric field has magnitude Ej = 426 V/m, and the angle between the electric field and the surface of the cube is 0 = 15 degrees. On the bottom of the cube, the electric field has the same magnitude Ej, and the angle between the electric field and the surface of the cube is also 0 = 15 degrees. On the top of the cube and the left side of the cube, the electric field is zero. On half of the front and back faces, the electric field has magnitude Ej and is parallel to the face; on the other half of the front and back faces, the electric field is zero. One edge of the cube is 41 cm long. Part 1 What is the net electric flux on this cubical surface? Net electric flux = i V•m Save for Later Attempts: 0 of 4 used Submit Answerarrow_forward
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