Answered: Consider a closed triangular box…

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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Consider a closed triangular box resting within a horizontal electric field of magnitude E = 7.02 × 10ª N/C as shown in the figure below.
30.0 cm
60.0°
10.0 cm
(a) Calculate the electric flux through the vertical rectangular surface of the box.
kN · m2/c
(b) Calculate the electric flux through the slanted surface of the box.
kN · m2/c
(c) Calculate the electric flux through the entire surface of the box.
kN · m2/c

Expert Solution

Step 1

(given)

Electric field (E) in magnitude = 7.02 $\times 10^{4} N / C$

(a)

Electric flux through the vertical rectangular surface of the box:

$ϕ = E A \cos θ = E A \cos 0 ° = E A = - (7.02 \times 10^{4} N / C) \times (0.30 m \times 0.10 m) \times (1) = - 2106 N m^{2} / C = - 2.106 k N m^{2} / C$

Answer:

Electric flux through the vertical rectangular surface of the box = - 2.106 kN m²/C

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