| A square surface of side L m is in the plane of the paper. A uniform electric field \[\vec{E}\] (V/m), also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is: [AIPMT (S) 2006] |
 |
A) \[E{{L}^{2}}/(2{{\varepsilon }_{0}})\]
B) \[E{{L}^{2}}/2\]
C) zero
D) \[E{{L}^{2}}\]
Correct Answer: C
Solution :
| [c] Electric flux \[({{\phi }_{e}})\] is a measure of the number of field lines crossing a surface, The number of field lines passing through unit area (N/S) will be proportional to the electric field, or, |
| \[\frac{N}{S}\propto \,\,E\Rightarrow \,\,N\,\propto \,ES\] |
| The quantity ES is the electric flux through surfaces S. |
| As we have seen in the problem that, lines of force that enter the closed surface leave the surface immediately, so no electric flux is bound to the system. Hence, electric flux is zero. |
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