If an electric field is given by $\vec{E} = 10 \hat{i} + 3 \hat{j} + 4 \hat{k}$,calculate the electric flux through a surface of area $A = 10 \text{ units}$ lying in the $yz$-plane.

Assertion : Electric lines of force never cross each other.
Reason : Electric field at a point superimpose to give one resultant electric field.

Electric charge is uniformly distributed along a long straight wire of radius $1\, mm$. The charge per $cm$ length of the wire is $Q$ $coulomb$. Another cylindrical surface of radius $50\, cm$ and length $1\, m$ symmetrically encloses the wire as shown in the figure. The total electric flux passing through the cylindrical surface is

Five charges $+q, +5q, -2q, +3q$ and $-4q$ are situated as shown in the figure. The electric flux due to this configuration through the surface $S$ is

$A$ circular disc of radius $R$ carries surface charge density $\sigma(r) = \sigma_0 \left(1 - \frac{r}{R}\right)$,where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is:

$A$ hollow cylinder has a charge $q$ at its center. If $\phi$ is the electric flux associated with the curved surface $B,$ the flux linked with the plane surface $A$ will be $:-$