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A metal sphere with radius ra = 1.20 cm is supported on an

University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman ISBN: 9780321675460 31

Solution for problem 48E Chapter 23

University Physics | 13th Edition

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University Physics | 13th Edition | ISBN: 9780321675460 | Authors: Hugh D. Young, Roger A. Freedman

University Physics | 13th Edition

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Problem 48E

A metal sphere with radius ra = 1.20 cm is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb = 9.60 cm. Charge +q is put on the inner sphere and charge -q on the outer spherical shell. The magnitude of q is chosen to make the potential difference between the spheres 500 V, with the inner sphere at higher potential. (a) Use the result of Exercise 23.41(b) to calculate q. (b) With the help of the result of Exercise 23.41(a), sketch the equipotential surfaces that correspond to 500, 400, 300, 200, 100, and 0 V. (c) In your sketch, show the electric field lines. Are the electric field lines and equipotential surfaces mutually perpendicular? Are the equipotential surfaces closer together when the magnitude of is largest? 23.41 .. CALC A metal sphere with radius ra is supported on an insulating stand at the center of a hollow, metal, spherical shell with radius rb. There is charge +q on the inner sphere and charge -q on the outer spherical shell. (a) Calculate the potential V(r) for (i) r < ra; (ii) ra < r < rb; (iii) r > rb. (Hint: The net potential is the sum of the potentials due to the individual spheres.) Take V to be zero when r is infinite. (b) Show that the potential of the inner sphere with respect to the outer is (c) Use Eq. (23.23) and the result from part (a) to show that the electric field at any point between the spheres has magnitude (d) Use Eq. (23.23) and the result from part (a) to find the electric field at a point outside the larger sphere at a distance r from the center, where r > rb. (e) Suppose the charge on the outer sphere is not -q but a negative charge of different magnitude, say -Q. Show that the answers for parts (b) and (c) are the same as before but the answer for part (d) is different.

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PHYS 2401 WEEK 4 EX. Look at +2Q plate by itself Look at ­Q by itself Now look at them together Now add the Es for both positive and negative plates for each section I Section (­Q/( *A)) + (Q/(2* *A)) = (­Q/(2* *A)) II Section (Q/( *A)) + (Q/(2* *A)) = (3Q/(2* *A)) III Section (Q/( *A)) + (­Q/(2* *A)) = (Q/(2* *A)) Note: E = σ / So, σ = E* Positive Plate Left side: (Q/(2A)) Right side: (3Q/(2A)) + (4Q/(2A)) = 2(Q/A) Negative Plate Left side: (­3Q/(2A)) Right side: (Q/(2A)) + (­Q/(2A)) CH 25: Electrical Potential Work Energy Theorem W cons conservative work W : non conservative work ext W :nett w

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Chapter 23, Problem 48E is Solved
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Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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A metal sphere with radius ra = 1.20 cm is supported on an