4) A ring of radius a carries a nonuniform charge density of 1 = 1, sin o. Let the ring lie in the x-y plane and place the center of the circle at the origin. Let o be the angle around the ring in the x-y plane. Calculate the electric field at a point Pa distance z above the ring along the z-axis. What is the direction of the field at the point P if an angle of o lines up with the +x axis?

Physics for Scientists and Engineers
10th Edition
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Chapter23: Continuous Charge Distributions And Gauss's Law
Section: Chapter Questions
Problem 5P: Example 23.3 derives the exact expression for the electric field at a point on the axis of a...
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Answers to help answer question in picture. Need help with question 4 (alone in seperate picture).

4) A ring of radius a carries a nonuniform charge density of 1 = 1, sin o. Let the ring lie in the
x-y plane and place the center of the circle at the origin. Let o be the angle around the ring in the
x-y plane. Calculate the electric field at a point Pa distance z above the ring along the z-axis.
What is the direction of the field at the point P if an angle of o lines up with the +x axis?
Transcribed Image Text:4) A ring of radius a carries a nonuniform charge density of 1 = 1, sin o. Let the ring lie in the x-y plane and place the center of the circle at the origin. Let o be the angle around the ring in the x-y plane. Calculate the electric field at a point Pa distance z above the ring along the z-axis. What is the direction of the field at the point P if an angle of o lines up with the +x axis?
Solution
(C)
E = KQx
k = L
9 x10 9 w mtIce
= 2.25m
x²+q? = (Iis)?+(0.04)?
父 =1.5
,a = 0. O4
で)
E =
9 X 1o 9 X 1.5 Yux 10-)
521 x 109
(2.25 J3e
3. 3 8
E = I5.97X10C
こ(6
X 10
NIC
6 y.8 X 104
11.1 x lo4
at
父 = 1.8 m
*へ
E =
5.832
at x =
2.1 m ;
75.6 X 104
8. 15 X104NIC
ニ
9.2 7
at
= 2. u m E
86.4 X1
6.26 x lo° C
ニ
ニ
13.8
at
4.3 3 x 10" NIc
X = 2.7 m,
E =
97.2 x loY
ニ
19.7
from
gu aph we Ceeleulatke that
as dis tem ce of Pointx
E入o!s
from center of eriny the
12
Value of
electric field decrease.
る
3
2.5
3.5 u
Transcribed Image Text:Solution (C) E = KQx k = L 9 x10 9 w mtIce = 2.25m x²+q? = (Iis)?+(0.04)? 父 =1.5 ,a = 0. O4 で) E = 9 X 1o 9 X 1.5 Yux 10-) 521 x 109 (2.25 J3e 3. 3 8 E = I5.97X10C こ(6 X 10 NIC 6 y.8 X 104 11.1 x lo4 at 父 = 1.8 m *へ E = 5.832 at x = 2.1 m ; 75.6 X 104 8. 15 X104NIC ニ 9.2 7 at = 2. u m E 86.4 X1 6.26 x lo° C ニ ニ 13.8 at 4.3 3 x 10" NIc X = 2.7 m, E = 97.2 x loY ニ 19.7 from gu aph we Ceeleulatke that as dis tem ce of Pointx E入o!s from center of eriny the 12 Value of electric field decrease. る 3 2.5 3.5 u
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