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

Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter24: Gauss’s Law
Section: Chapter Questions
Problem 24.63CP: A dosed surface with dimensions a = b= 0.400 111 and c = 0.600 in is located as shown in Figure...
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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
Transcribed Image Text: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 ×104 N/C

(a)

Electric flux through the vertical rectangular surface of the box:

ϕ=EA cos θ  =EA cos 0°  =EA  =-7.02×104N/C×0.30 m × 0.10 m×1  =-2106 N m2/C  =-2.106 kN m2/C

Answer:

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

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