Density, density, density. (a) A charge -328e is uniformly distributed along a circular arc of radius 6.00 cm, which subtends an angle of 72°. What is the linear charge density along the arc? (b) A charge -328e isuniformly distributed over one face of a circular disk of radius 3.50 cm. What is the surface charge density over that face? (c) A charge -328e is uniformly distributed over the surface of a sphere of radius 2.00 cm. Whatis the surface charge density over that surface? (d) A charge -328e is uniformly spread through the volume of a sphere of radius 3.30 cm. What is the volume charge density in that sphere?(a) NumberUnits(b) NumberUnits(c) NumberUnits(d) NumberUnits

Answered: Density, density, density. (a) A charge…

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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Density, density, density. (a) A charge -328e is uniformly distributed along a circular arc of radius 6.00 cm, which subtends an angle of 72°. What is the linear charge density along the arc? (b) A charge -328e is
uniformly distributed over one face of a circular disk of radius 3.50 cm. What is the surface charge density over that face? (c) A charge -328e is uniformly distributed over the surface of a sphere of radius 2.00 cm. What
is the surface charge density over that surface? (d) A charge -328e is uniformly spread through the volume of a sphere of radius 3.30 cm. What is the volume charge density in that sphere?
(a) Number
Units
(b) Number
Units
(c) Number
Units
(d) Number
Units

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Hello. Since your question has multiple sub-parts, we will solve the first three sub-parts (a, b, and c) for you. If you want the remaining sub-parts to be solved, then please resubmit the whole question and specify those sub-parts you want us to solve.

(a)

The linear charge density along the arc can be determined as follows.

$\begin{array}{rcl} λ & = & \frac{q}{L} \\ = & \frac{q}{r θ} \\ = & \frac{- 328 e}{(6.00 cm) (72 °)} \\ = & \frac{- 328 (- 1.6 \times 10^{- 19} C)}{(6.00 \times 10^{- 2} m) (72 ° (\frac{π rad}{180 °}))} \\ = & 6.96 \times 10^{- 16} C / m \end{array}$

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