A solid non-conducting sphere has a non-uniform charge distribution of p = ar, where r is the radial distance from the sphere's center and a is aconstant. If the sphere has a radius of 6 cm and the magnitude of the E field at r = 3 cm from the center is 5.6 N/C, what is the value of a?Assume the charge is positive and the E field points radially outward. You can state your answer in units of C/m4 or µC/m“. Remember that dV fora sphere is 47r²dr.

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Answered: A solid non-conducting sphere has a…

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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Two large, nonconducting plates are suspended 2.83 cm apart. Plate 1 has an area charge density of +92.3 µC/m², and plate 2 has an area charge density of +10.1 µC/m². Treat each plate as an infinite sheet. How much electrostatic energy UE is stored in 2.29 cm³ of the space in region A? UE = What volume V of the space in region B stores an equal amount of energy? V = J 3 cm³ Plate 1 Region A Region B Plate 2 Region C
An infinite sheet of charge with surface charge density o = -87 C/m? lives in the x-y plane. A thin rod of charge with length L = 13.0 cm is placed along the +z-axis that the end closest to the sheet is located a distance d = 2.0cm away from the sheet. When held at this position, the linear charge density of the rod can be described by K =az, where a = +35 nC/m. 1, What is the force F on the rod from the sheet? 2, Can you draw a picture to represent it?
Two infinite nonconducting plates are parallel to each other, with a distance d = 12.0 cm between them. Each plate carries a uniform charge distribution with the surface charge density o = 4.28 μ C/m². What is the magnitude of the electric field at point R (with r = 17.0 cm)? ச 16 R 22 0.00 N/C 4.84 x 105 N/C O 5.80 x 105 N/C 3.87 x 105 N/C 3.39 X 105 N/C X
A hollow conducting sphere has an inside radius of r1 = 0.17 m and an exterior radius of r2 = 0.27 m. The sphere has a net charge of Q = 5.4E-06 C. What is the field E1 in N/C 1 m from the sphere's exterior surface? What is the magnitude of the field at a distance of 0.05 m from the center of the sphere?
The figure below shows a small, charged bead, with a charge of q = +41.0 nC, that moves a distance of d = 0.174 m from point A to point B in the presence of a uniform electric field E of magnitude 255 N/C, pointing right. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) What is the magnitude (in N) and direction of the electric force on the bead? magnitude Ndirection (b) What is the work (in J) done on the bead by the electric force as it moves from A to B? J (c) What is the change of the electric potential energy (in J) as the bead moves from A to B? (The system consists of the bead and all its surroundings.) PEB − PEA = J (d) What is the potential difference (in V) between A and B? VB − VA = V
A circular plastic disk with radius R = 2.00 cm has a uniformly distributed charge Q = +(2.00 x 106)e on one face. A circular ring of width 30 µm is centered on that face, with the center of that width at radius r = 0.50 cm. In coulombs, what charge is contained within the width of the ring?
Early in the 20th century, a leading model of the structure of the atom was that of English physicist J. J. Thomson (the discoverer of the electron). In Thomson’s model, an atom consisted of a sphere of positively charged material in which were embedded negatively charged electrons, like chocolate chips in a ball of cookie dough. Consider such an atom consisting of one electron with mass m and charge -e, which may be regarded as a point charge, and a uniformly charged sphere of charge +e and radius R. By that time time, it was known that excited atoms emit light waves of only certain frequencies. In his model, the frequency of emitted light is the same as the oscillation frequency of the electron (s) problems in the atom. What radius (in millimeter) would a Thomson-model atom need for it to produce red light of frequency 4.57 x 1014 Hz? (Don't express your answer in scientific notation)
Two spherical, hollow conductors are concentrically nested as shown in the cross-sectional diagram below and electrically isolated from each other. A net charge of -3 nC is divided between the conductors, with a total of -12 nC on the inner conductor and +9 nC on the outer one. The charges are, of course, free to move between the surfaces within each shell but cannot move from one shell to the other. The inner conductor has an inner radius of a=2 cm, outer radius of b=3 cm. The outer conductor has an inner radius of c=6 cm, an outer radius of d=8 cm. a (a) In equilibrium, all of the excess charge is found on the surfaces. Why is there no charge between a and b, nor between c and d? (b) Determine the net charge on each surface (ie how much charge is there at r = a, at r = b, at r = c and at r= = d). (c) Is the charge on the outer surface equal to the net charge on the outer container, the net charge of the system, or some other amount? Explain briefly. (d) Find the surface charge…
The figure below shows a small, charged bead, with a charge of q = +42.0 nC, that moves a distance of d = 0.189 m from point A to point B in the presence of a uniform electric field E of magnitude 270 N/C, pointing right. A positive point charge q is initially at point A, then moves a distance d to the right to point B. Electric field vector E points to the right. (a) What is the magnitude (in N) and direction of the electric force on the bead? magnitude Ndirection (b) What is the work (in J) done on the bead by the electric force as it moves from A to B? J (c) What is the change of the electric potential energy (in J) as the bead moves from A to B? (The system consists of the bead and all its surroundings.) PEB − PEA = J (d) What is the potential difference (in V) between A and B? VB − VA = V

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