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O webwork.phys.yorku.ca/webwork2/PHYS1010/w (1 point) . THE PROBLEM A laboratory room is filled with a constant electric field which enters through the left wall and exists through the right wall. An empty wooden cube sits on the floor of the lab. The field has magnitude E-1.71 kN/C and the cube has dimensions 1x L × L where L-122 cm. Calculate the electric flux through each of the six sides on the wooden cube when (a) its left side faces the left wall of the lab, its right side faces the right wall, front side faces front wall, back side faces back wall, top side faces ceiling. and bottom side faces floor (b) it is rotated by 45 degrees, so its left side faces the labs left-back corner, its right side faces the labs right-front corner, its front side faces the labs left-front corner, its back side faces the labs right- back corner, top side faces ceiling, and bottom side faces floor PAPER SOLUTION INTERPRET DEVELOP Derive an algebraic expression for the electric flux through each side in part (a). ELA2 rght ELA2 efs

Derive an algebraic expression for the electric flux through each side in part (a). EL 2 right EL 2 front = = EL 2cos(pl/2) EL 2cos(pi/2) top EL 2cos(pl/2) 重bottom EL 2cos(pi/2) Derive an algebraic expression for the electric flux through each side in part (b). left = ELA2cos(5pi4) right = EL 2cos(pl/4) ELA2cos(Spi/4) back = EL 2cos pi 4) ELA2cos(pi 2) EL 2cos(pi/2) EVALUATE ㄥ

le ft 2.54E03 /С Nm 2.54E03 /С Frone Nm Nm /С Pack Nm /С Nm /С Nm /C Calculate a numerical value for the electric flux through each side in part (b) 2.54E03 /С rght 2.54E03 /С Nm 2.53E03 Nm /C 2.54E03 Nm

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Answer #1

The electric flux magnitude through a surface is given by the products of electric field,area of the surface and the cosine of angle between the electric field and the area vector. By use of this formula we can solve the given problem.

Feig に122 cm 2 .,. 귀지03 싶1.31X103 て ( 122x10-2m)2. augus。 Cus ( 135) ㄈㄧ-1729.703 tin?) = 799.703

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