(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 .

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN: 9781133104261

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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Principles of Physics: A Calculus-Based Text

Ch. 19.2 - Three objects are brought close to one another,...Ch. 19.3 - Three objects are brought close to one another,...Ch. 19.4 - Object A has a charge of +2 C, and object B has a...Ch. 19.5 - A test charge of +3 C is at a point P where an...Ch. 19.6 - Rank the magnitudes of the electric field at...Ch. 19.9 - If the net flux through a gaussian surface is...Ch. 19.9 - Consider the charge distribution shown in Active...Ch. 19 - A point charge of 4.00 nC is located at (0, 1.00)...Ch. 19 - Charges of 3.00 nC, 2.00 nC, 7.00 nC, and 1.00 nC...Ch. 19 - An object with negative charge is placed in a...

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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.