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Consider the electric dipole of Example 21.14. (a) Derive

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

Solution for problem 60E Chapter 21

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 60E

Consider the electric dipole of Example 21.14. (a) Derive an expression for the magnitude of the electric field produced by the dipole at a point on the -axis in Fig. 21.33. What is the direction of this electric field? (b) How does the electric field at points on the -axis depend on when is very large?

Step-by-Step Solution:

Solution 60E Introduction First we have to calculate the electric field due to dipole at its axis. Then we have to find out how the electric field varies at very large distance from the dipole moment. Step 1: Let us consider a point P at the axis of the dipole as shown in the above figure. We have to calculate the electric field at P at a distance x from the center of the dipole. To calculate the electric field due to the dipole, we have to calculate the electric field due to both the positive charge and the negative charge. Now since the point is on the axis, the distance from the charges to the point is same. Now if the separation between the charge is d, then the distance from the charge to the point is r = (d/2) + x2 Now the magnitude of the electric field for both the positive charge and negative charge will be q E = 2 0 40[(d/2) +x ] Now the directions of the electric field will be as shown in the figure. The E represents + the electric field due to positive charge and E represents the electric field due to negative charge. The horizontal component will be equal and opposite to each other and will cancel out, only the vertical component will survive, which will be in the downward direction. Now the vertical component is given by E V E sin0 Where = tan 1d/2 x Hence the vertical component is given by 1 d/2 q 1 d/2 E V E sin0tan x ) = 2 2 sin(tan x ) 40[(d/2) +x ] Hence the total downward electric field is 2q E = 2E = V 2 2 sin(tan 1 d/2) 40[(d/2) +x ] x Hence the magnitude of the electric field due to dielectric is 2q sin(tan 1 d/2) 40[(d/2) +x ] x And will be directed downwards.

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Chapter 21, Problem 60E is Solved
Textbook: University Physics
Edition: 13
Author: Hugh D. Young, Roger A. Freedman
ISBN: 9780321675460

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Consider the electric dipole of Example 21.14. (a) Derive