Question

An electric field of intensity 3.7 kN/C is applied along the x-axis. Calculate the electric flux through a rectangular plane 0.350 m wide and 0.700 m long if the following conditions are true:
a. The plane is parallel to the yz-plane.
___________N·m2/C
b. The plane is parallel to the xy-plane.
___________N·m2/C
c. The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.
___________N·m2/C

Answer 1

Answer:

a) $906.5\ N .m^2/C$

b) $0\ N . m^2/C$

c) $742.5\ N . m^2/C$

Explanation:

Area of the plane is,

a) The plane is parallel to the yz-plane.

$A = (0.350)(0.700)\\= 0.245\ m^2\\\phi_{E} = EA \cos \theta \\= (3.70 \times 10^3)(0.245) \cos0^{\circ}\\\phi_{E} = 906.5\ Nm^2/C$

b) The plane is parallel to the x-axis, the normal line of the area in at a right angle

$\theta = 90^{\circ}\\ \phi_{E} = EA \cos \theta\\ \phi_{E} = (3.70 \times 10^3)(0.245) \cos 90^{\circ}\\ \phi_{E} = 0\ N . m^2/C$

c) The plane contains the y-axis, and its normal makes an angle of $35.0^{\circ}$ with the x-axis $35.0^{\circ}$

$\phi_{E} = EA \cos \theta\\ \phi_{E} = (3.70 \times 10^3)(0.245) \cos 35^{\circ}\\ \phi_{E} = 742.5\ N . m^2/C$

Answer 2

Answer:

a)906.5 Nm^2/C

b) 0

c) 742.56132 N•m^2/C

Explanation:

a) The plane is parallel to the yz-plane.

We know that

flux ∅= EAcosθ

3.7×1000×0.350×0.700=906.5 N•m^2/C

(b) The plane is parallel to the xy-plane.

here theta = 90 degree

therefore,

0  N•m^2/C

(c) The plane contains the y-axis, and its normal makes an angle of 35.0° with the x-axis.

therefore, applying the flux formula we get

3.7×1000×0.3500×0.700×cos35°= 742.56132 N•m^2/C