The total electric fleld at point P, a distance x from the center of a ring of radius a, is given by the following equation. Substituting the given values for the charge and radius, we have the following. kxQ E = (x² + a?y3/2 (8.99 x 10° N. m2/c2)( 77.5 V 77.5 x 10-6 cx 3/2 0.100 0.1 m)) 6.96725 6.97 x 105 N - m²/cx (x2 + (0.100 0.1 m)) Step 4 Using a calculator, we substitute the distances for x and evaluate the electric field. (a) At x = 0.0100 m, E x 106 I N/C. %3D (b) At x = 0.0500 m, E = x 10 i N/C. (c) At x = 0.300 m, E x 106 i N/C. (d) At x = 1.00 m, E x 105 I N/C.

Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter24: Gauss’s Law
Section: Chapter Questions
Problem 24.3OQ: In which of the following contexts ran Gausss law not be readily applied to find the electric field?...
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The total electric field at point P, a distance x from the center of a ring of radius a, is given by the following equation. Substituting the given values for the charge and radius, we have the following.
Step 3
The total electric fleld at point P, a distance x from the center of a ring of radius a, is given by the following equation. Substituting the given values for the charge and radius, we have the following.
keXQ
E =
(x² + a²j3/2
(8.99 x 10° N· m²/c2)( 77.5
77.5 x 10-6 c)x
0.100
0.1 m
6.96725
6.97
x 105 N - m?/c)x
(x2 +
0.100
0.1 m))"
Step 4
Using a calculator, we substitute the distances for x and evaluate the electric field.
(a) At x = 0.0100 m, E
x 106 i N/C.
(b) At x = 0.0500 m, E
x 107 î N/C.
%3D
(c) At x = 0.300 m, E =
x 106 i N/C.
(d) At x = 1.00 m, E =
x 105 i N/C.
Transcribed Image Text:Step 3 The total electric fleld at point P, a distance x from the center of a ring of radius a, is given by the following equation. Substituting the given values for the charge and radius, we have the following. keXQ E = (x² + a²j3/2 (8.99 x 10° N· m²/c2)( 77.5 77.5 x 10-6 c)x 0.100 0.1 m 6.96725 6.97 x 105 N - m?/c)x (x2 + 0.100 0.1 m))" Step 4 Using a calculator, we substitute the distances for x and evaluate the electric field. (a) At x = 0.0100 m, E x 106 i N/C. (b) At x = 0.0500 m, E x 107 î N/C. %3D (c) At x = 0.300 m, E = x 106 i N/C. (d) At x = 1.00 m, E = x 105 i N/C.
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