The electric flux from a cube of edge $l$ is $\phi$. If an edge of the cube is made $2l$ and the charge enclosed is halved,its value will be

Which statement$(s)$ among the following are incorrect:
$(i)$ $A$ negative test charge experiences a force opposite to the direction of the field.
$(ii)$ The tangent drawn to a line of force represents the direction of electric field.
$(iii)$ The electric field lines never intersect.
$(iv)$ The electric field lines form a closed loop.

If the electric flux entering and leaving an enclosed surface is $\phi_1$ and $\phi_2$ respectively, then the charge enclosed in the surface is ($\varepsilon_0 =$ permittivity of free space).

$A$ point charge of $10^{-7} \text{ C}$ is situated at the centre of a cube of $1 \text{ m}$ side. The electric flux through its surface is

An infinitely long uniform line charge distribution of charge per unit length $\lambda$ lies parallel to the $y$-axis in the $y-z$ plane at $z=\frac{\sqrt{3}}{2} a$ (see figure). If the magnitude of the flux of the electric field through the rectangular surface $A B C D$ lying in the $x-y$ plane with its center at the origin is $\frac{\lambda L }{ n \varepsilon_0}$ (where $\varepsilon_0$ is the permittivity of free space),then the value of $n$ is

$A$ square surface of side $L \; m$ in the plane of the paper is placed in a uniform electric field $E \; (V/m)$ acting along the same plane at an angle $\theta$ with the horizontal side of the square as shown in the figure. The electric flux linked to the surface,in units of $V \cdot m$,is: