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Physics
A point charge + q is placed at the mid point of a cube of side L . The electric flux emerging from the cube is :
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Q. A point charge $ + q $ is placed at the mid point of a cube of side $ L $ . The electric flux emerging from the cube is :
UPSEE
UPSEE 2005
A
$ \frac{q}{\varepsilon_{0}} $
B
$ \frac{q}{6L^2\varepsilon_{0}} $
C
$ \frac{6qL^{2}}{\varepsilon_{0}} $
D
zero
Solution:
From Gauss theorem, the electric flux $(\phi)$ through any closed surface is equal to $\frac{1}{\varepsilon_{0}}$ times the net charge $q$ enclosed by the surface. Thus,
$\phi=\int \vec{ E } \cdot d \vec{ A }=\frac{q}{\varepsilon_{0}}$