Draw the electric field lines for a positive point charge.

$A$ uniformly charged conducting sphere of diameter $14 \ cm$ has a surface charge density of $40 \ \mu C/m^2$. The total electric flux leaving the surface of the sphere is nearly (Permittivity of free space $\varepsilon_0 = 8.85 \times 10^{-12} \ C^2/N \cdot m^2$)

$A$ charge $Q$ is placed at the center of a closed cube. The flux passing through any one face of the cube will be .......

If electric flux through a cubic Gaussian surface is $1.9 \times 10^5 \text{ Nm}^2 \text{C}^{-1}$,then the electric charge at its center is . . . . . . . (Edge length of cube = $9.0 \text{ cm}$).

If $\oint_s \vec{E} \cdot \overrightarrow{d S}=0$ over a surface,then:

$A$ point charge $+10 \; \mu C$ is at a distance $5 \; cm$ directly above the centre of a square of side $10 \; cm$,as shown in the figure. What is the magnitude of the electric flux through the square?