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 ${\varepsilon _0} = 8.85 \times {10^{ – 12}}\,{C^2}/N – {m^2},{R_E} = 6.37 \times {10^6}\,m$]

An ellipsoidal cavity is carved within a perfect conductor. A positive charge $q$ is placed at the centre of the cavity. The points $A$ and $B$ are on the cavity surface as shown in the figure. Then

A charged shell of radius $R$ carries a total charge $Q$. Given $\Phi$ as the flux of electric field through a closed cylindrical surface of height $h$, radius $r$ and with its center same as that of the shell. Here, center of the cylinder is a point on the axis of the cylinder which is equidistant from its top and bottom surfaces. Which of the following option(s) is/are correct ? $\epsilon_0$ is the permittivity of free space]

$(1)$ If $h >2 R$ and $r > R$ then $\Phi=\frac{ Q }{\epsilon_0}$

$(2)$ If $h <\frac{8 R }{5}$ and $r =\frac{3 R }{5}$ then $\Phi=0$

$(3)$ If $h >2 R$ and $r =\frac{4 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$

$(4)$ If $h >2 R$ and $r =\frac{3 K }{5}$ then $\Phi=\frac{ Q }{5 \epsilon_0}$

If the electric field intensity in a fair weather atmosphere is $100 \,V / m$, then the total charge on the earth's surface is ………… $C$ (radius of the earth is $6400\,km$ )

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