If the electric field is given by $(5 \hat{i}+4 \hat{j}+9 \hat{k})$. The electric flux through a surface of area $20$ units lying in the $Y-Z$ plane will be (in units)

The electric flux for Gaussian surface A that enclose the charged particles in free space is (given $q_1$ = $-14\, nC$, $q_2$ = $78.85\, nC$, $q_3$ = $-56 \,nC$)

A few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $\mathrm{x}$-axis are shown in the figure. These lines suggest that $Image$

$(A)$ $\left|Q_1\right|>\left|Q_2\right|$

$(B)$ $\left|Q_1\right|<\left|Q_2\right|$

$(C)$ at a finite distance to the left of $\mathrm{Q}_1$ the electric field is zero

$(D)$ at a finite distance to the right of $\mathrm{Q}_2$ the electric field is zero

The magnitude of the average electric field normally present in the atmosphere just above the surface of the Earth is about $150\, N/C$, directed inward towards the center of the Earth . This gives the total net surface charge carried by the Earth to be......$kC$ [Given ${\varepsilon _0} = 8.85 \times {10^{ - 12}}\,{C^2}/N - {m^2},{R_E} = 6.37 \times {10^6}\,m$]

Explain the electric field lines and the magnitude of electric field.

What is the direction of electric field intensity ?