question_answer

Consider a region bounded by conical surface as shown in figure   In this region, E is in vertical upward direction. Find flux crossing this surface.

A)  \[E\pi {{r}^{2}}\,sin \theta \]               

B)  \[\frac{E}{3}\pi r/\,sin \theta \]

C)  \[E\pi {{r}^{2}}\]                   

D)  \[\frac{E}{3}\pi {{r}^{2}}/\,cos \theta \]

Correct Answer: C

Solution :

Construct a plane surface on the base of cone as shown in figure as charge enclosed by this closed surface is zero, so flux crossing this surface is equal to zero. i.e.,       \[\phi =0\] \[\Rightarrow \,\,\,{{\phi }_{plane\,\,surface}}+\,\,{{\phi }_{plane\,\,surface}}\,\,=\,\,0\] \[{{\phi }_{curved\,\,surface}}~=\,\,-{{\phi }_{plane\,\,surface}}\] \[=\,\,-\left( E.A \right)\] \[=\,\,-EA cos 180{}^\circ \] \[=\,\,+E\pi \,{{r}^{2}}\,\,=\,\,E\pi \,{{r}^{2}}\]