(a) Explanation Figure 1 represents electric field at a point P due to the elemental length d x . Write the expression for electric field at a point P due to elemental length d x . d E = k e d q x 2 + y 2 (I) The direction of the electric field at point P is along the line joining the element to point P . By symmetry the electric field at point P along horizontal direction is zero E x = 0 . The net electric field at a point P is only due to the vertical component, which is given by E y = ∫ d E y (II) From Figure 1, the above equation is reduced to E = ∫ d E cos θ From Figure 1 cos θ is equal to y x 2 + y 2 . Thus the above equation is reduced to E = ∫ d E y x 2 + y 2 (III) Use equation (I) in (III) (b) To determine To show that the electric field of a rod of infinite length is E = 2 k e λ / y .

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

5th Edition

ISBN: 9781133422013

Author: Raymond A. Serway; John W. Jewett

Publisher: Cengage Learning

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Chapter 19, Problem 18P

(a)

To determine

To show that the electric field at a point P is E=2keλsinθ0y.

(b)

To determine

To show that the electric field of a rod of infinite length is E=2keλ/y.

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Chapter 19, Problem 17P

Chapter 19, Problem 19P

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Chapter 19 Solutions

Bundle: Principles of Physics: A Calculus-Based Text, 5th + WebAssign Printed Access Card for Serway/Jewett's Principles of Physics: A Calculus-Based Text, 5th Edition, Multi-Term

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To show that the electric field at a point P is E = 2 k e λsin θ 0 y .

Solution Summary: The author illustrates the electric field at a point P due to the elemental length dx.