PRACTICE: Consider the setup shown in the figure below, where the arc is a semicircle withradius r. The total charge Q is negative, and distributed uniformly on the semicircle. Thecharge on a small segment with angle Ae is labeled Aq. The magnitude of the x-component ofthe electric field at the center, due to Aq, is given by!!IIIII IV1. AE, = k|Aq| (cos6)r6. AE, =kSyi sin e2. AE, =7. AE; = k|A| (sin@) r3. AE, = 1:|Ayl (cos @) rk: JAg| sin e8. AE, =4. AE, = kAq} (sin #) rkAy cos e9. JE, =5.AE, = A cos #10. AE, = k|Aylr%3DIf correct plese expleis.

Answered: Image /qna-images/answer/6bca90f1-ca2c-4b20-8957-9902fcd7717c.jpg

PRACTICE: Consider the setup shown in the figure below, where the arc is a semicircle with radius r. The total charge Q is negative, and distributed uniformly on the semicircle. The charge on a small segment with angle A0 is labeled Aq. The magnitude of the x-component of the electric field at the center, due to Aq, is given by !! II III IV 1. AE, = k|Sq| (cos@) r よ 6. AE, = 2. AE, = kSyi sin e 7. AE, = kSy|l (sin@) r 3. AE, = 1:|A4| (cos@) r 8. ΔΕ, = 4. AE, = k|Aq! (sin@) r kAq| cos e 9. JE, = 5.JE, = 10. JE, = k|yr If correct plese explein,

PRACTICE: Consider the setup shown in the figure below, where the arc is a semicircle with radius r. The total charge Q is negative, and distributed uniformly on the semicircle. The charge on a small segment with angle A0 is labeled Aq. The magnitude of the x-component of the electric field at the center, due to Aq, is given by !! II III IV 1. AE, = k|Sq| (cos@) r よ 6. AE, = 2. AE, = kSyi sin e 7. AE, = kSy|l (sin@) r 3. AE, = 1:|A4| (cos@) r 8. ΔΕ, = 4. AE, = k|Aq! (sin@) r kAq| cos e 9. JE, = 5.JE, = 10. JE, = k|yr If correct plese explein,

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

PRACTICE: Consider the setup shown in the figure below, where the arc is a semicircle with
radius r. The total charge Q is negative, and distributed uniformly on the semicircle. The
charge on a small segment with angle Ae is labeled Aq. The magnitude of the x-component of
the electric field at the center, due to Aq, is given by
!!
II
III IV
1. AE, = k|Aq| (cos6)r
6. AE, =
kSyi sin e
2. AE, =
7. AE; = k|A| (sin@) r
3. AE, = 1:|Ayl (cos @) r
k: JAg| sin e
8. AE, =
4. AE, = kAq} (sin #) r
kAy cos e
9. JE, =
5.AE, = A cos #
10. AE, = k|Aylr
%3D
If correct plese expleis.

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Six negative charges and two positive charges, all of equal magnitude Q, are arranged as shown in the Fig. A. The radius of the circle is r. The magnitude of the electric field generated by a single charge Qis, E = 150 N/C at the origin. (a) What is the magnitude of the net electric field at the origin O. (b) What is the direction of the net electric field? Choose from the arrows shown in Fig. B. -Q D* 0 Fig. B -Q' Fig. A Your Answer: (a) Magnitude: (b) Direction: (Choose from the arrow).
A charge distribution creates the following electric field throughout all space: E(r, 0, q) = (3/r) (r hat) + 2 sin cos sin 0(theta hat) + sin cos p (phi hat). Given this electric field, calculate the charge density at location (r, 0, p) = (ab.c).
My answer is 42.9 cm but it is wrong but the unit CM si correct
PLEASE ANSWER ALL PARTS ASAP!!!!!!!!
A thin rod is bent into a circular arc that subtends an angle 2θ of a circle centered at P. A total charge, Q, is distributed uniformly over the full rod. Let the positive x direction be towards the right of the page, and the positive y direction is towards the top of the page. In Cartesian unit-vector format, what is the electric field at P?