(a) Calculate the electric flux through the vertical rectangular surface of the

Question

(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

An electric field of magnitude 3.14 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 38.0° with the x axis.

Explanation / Answer

E = 7.36*10^4 N/C

electric flux phi= EAcos(theta)

(a) As the vertical rectangular surface is normal to the field theta = 0

phi = 7.36*10^4*(0.1*0.3)*cos(0) = 2208 N.m^2/c

As this flux is going in to the box it should be taken negative, hence the flux through the vertical rectangular surface is -2.21 kN.m^2/c

(b) As the slant surface is having the same vertical height and the width, the flux crossing it is same as the vertical rectangular surface. Or we can say that projection of the area vector for the slant surface normal to the field is equal to the area vector of the vertical rectangular surface hence the flux through it is same as that of the vertical surface but as it coming out of the box, taken positive and hence equal to +2.21 Nm^2/C

c) The entire surface of the box is a closed surface and as there is no charge within it, hence according to Gauss law the flux through the entire closed surface is zero.

E = 3.14 i kN/C , A = 0.35* 0.7 m^2

(a) phi = EA = 3.14*0.35*0.7

phi = 770 N.m^2/C

(b) phi =0

(c) phi = 3.14*0.35*0.7*cos(38)

phi = 606 N.m^2/C

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