If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface, then:
the electric field inside the surface is necessarily uniform.
the number of flux lines entering the surface must be equal to the number of flux lines leaving it.
the magnitude of electric field on the surface is constant.
all the charges must necessarily be inside the surface.
When electric flux is said to be positive, negative or zero ?
$(a)$ An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
$(b)$ Explain why two field lines never cross each other at any point?
Figure shows four charges $q_1, q_2, q_3$ and $q_4$ fixed in space. Then the total flux of electric field through a closed surface $S$, due to all charges $q_1, q_2, q_3$ and $q_4$ is
A charged particle $q$ is placed at the centre $O$ of cube of length $L$ $(A\,B\,C\,D\,E\,F\,G\,H)$. Another same charge $q$ is placed at a distance $L$ from $O$.Then the electric flux through $BGFC$ is
Let the electrostatic field $E$ at distance $r$ from a point charge $q$ not be an inverse square but instead an inverse cubic, e.g. $E =k \cdot \frac{q}{r^{3}} \hat{ r }$, here $k$ is a constant.
Consider the following two statements:
$(I)$ Flux through a spherical surface enclosing the charge is $\phi=q_{\text {enclosed }} / \varepsilon_{0}$.
$(II)$ A charge placed inside uniformly charged shell will experience a force.
Which of the above statements are valid?