Solution
(a)
We can see from the figure that on the left face E and DS are
parallel Therefore, the outward flux is
fL= E DS = – 200 ˆ ∆
i
S
i
=  + 200 DS, since  ˆ ∆
i
iS
= – DS
=  + 200 × p (0 05)2 = + 1