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Given electric field $ \overrightarrow {E} =[1 \times (d + x) \hat i +E_0 \hat j ] N/c$. If a closed surface as shown below is considered , find the change enclosed.


$(A)\;\frac{acd}{2} \in_0 \\ (B)\;\frac{abc \in _o}{2} \\ (C)\;\frac{abd \in _0}{2} \\ (D)\;None $

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$\phi ABCD=-acd$
$\phi CDEF=-bcE_0$
$\phi ABEF=bcE_0 + c \int \limits _o^a (d+x) dy$
$\qquad= bcE_0 +acd+ c \int \limits_0^a x dxy$
$\qquad= bcE_0 +acd +\large\frac{ca}{b} \int\limits_0^b x dx$
$\bigg[ \therefore \large\frac{x}{b}+ \frac{y}{a} $$=1=> \frac{dx}{b} =\frac{-dy}{a}\bigg]$
$\qquad= bcE_0 +acd +\large\frac{acb}{2}$
$\phi _{net}=\large\frac{q_{in}}{\in _0}$
$q_{in}=\large\frac{abc \in _o}{2}$
hence b is the correct answer.
answered Jan 15, 2014 by meena.p

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