Module 2 : Electrostatics
Lecture 7 : Electric Flux
 

Solution :
The outward normals to the triangular faces AED, BFC, as well as the normal to the base are perpendicular to $\vec E$. Hence the flux through each of these faces is zero. The vertical rectangular face ABFE has an area 0.06 m$^2$. The outward normal to this face is perpendicular to the electric field. The flux is entering through this face and is negative. Thus flux through ABFE is

\begin{displaymath}\phi_1 =- 0.06 \times 8\times 10^4 = -.48\times 10^4 \;\; {\rm N-m}^2/{\rm C}\end{displaymath}

To find the flux through the slanted face, we need the angle that the normal to this face makes with the horizontal electric field. Since the electric field is perpendicular to the side ABFE, this angle is equal to the angle between AE and AD, which is $\cos^{-1}(.3/.5)$. The area of the slanted face ABCD is 0.1 m$^2$. Thus the flux through ABCD is \begin{displaymath}\phi_2= 0.1\times 8\times 10^4\times (.3/.5) = +0.48\times 10^4 \;\;{\rm N-m}^2/ {\rm C}\end{displaymath}

The flux through the entire surface of the wedge is $\phi_1+\phi_2=0$

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