General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 31, Problem 2E
To determine
The argument that verify the relation between energy loss and number of atoms.
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Find the speed an alpha particle requires to
come within 3.9 × 10¬1ª m of a gold nucleus.
Coulomb's constant is 8.99 × 10º N · m² /C²,
the charge on an electron is 1.6 × 10-19
and the mass of the alpha particle is
6.64 × 10–27 kg.
Answer in units of m/s.
C,
Find the energy of the alpha particle.
Answer in units of MeV.
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 (
))
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 r, has charge -Q. The
electric field E at a radial distance r from the central axis is given by the function:
E = ae-T/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:
Voh =
Ed
Edr
Calculating the antiderivative or indefinite integral,
Vab = (-aaoe-r/a0 + 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/(
(erb/ao - eralao) + ß In(
) + bo (
))
Chapter 31 Solutions
General Physics, 2nd Edition
Ch. 31 - Prob. 1RQCh. 31 - Prob. 2RQCh. 31 - Prob. 3RQCh. 31 - Prob. 4RQCh. 31 - Prob. 5RQCh. 31 - Prob. 6RQCh. 31 - Prob. 7RQCh. 31 - Prob. 8RQCh. 31 - Prob. 9RQCh. 31 - Prob. 10RQ
Ch. 31 - Prob. 1ECh. 31 - Prob. 2ECh. 31 - Prob. 3ECh. 31 - Prob. 4ECh. 31 - Prob. 5ECh. 31 - Prob. 6ECh. 31 - Prob. 7ECh. 31 - Prob. 8ECh. 31 - Prob. 9ECh. 31 - Prob. 10ECh. 31 - Prob. 11ECh. 31 - Prob. 12ECh. 31 - Prob. 13ECh. 31 - Prob. 14ECh. 31 - Prob. 15ECh. 31 - Prob. 16ECh. 31 - Prob. 17ECh. 31 - Prob. 18ECh. 31 - Prob. 19ECh. 31 - Prob. 20ECh. 31 - Prob. 21ECh. 31 - Prob. 22ECh. 31 - Prob. 23ECh. 31 - Prob. 24ECh. 31 - Prob. 25ECh. 31 - Prob. 26ECh. 31 - Prob. 27ECh. 31 - Prob. 28ECh. 31 - Prob. 29ECh. 31 - Prob. 30ECh. 31 - Prob. 31ECh. 31 - Prob. 32ECh. 31 - Prob. 33ECh. 31 - Prob. 34ECh. 31 - Prob. 35ECh. 31 - Prob. 36ECh. 31 - Prob. 37ECh. 31 - Prob. 38ECh. 31 - Prob. 39ECh. 31 - Prob. 40ECh. 31 - Prob. 41ECh. 31 - Prob. 42ECh. 31 - Prob. 43ECh. 31 - Prob. 44ECh. 31 - Prob. 45ECh. 31 - Prob. 46ECh. 31 - Prob. 47ECh. 31 - Prob. 48ECh. 31 - Prob. 49ECh. 31 - Prob. 50ECh. 31 - Prob. 51ECh. 31 - Prob. 52ECh. 31 - Prob. 53ECh. 31 - Prob. 54ECh. 31 - Prob. 55ECh. 31 - Prob. 57E
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