$A$ charge $q$ is placed at the center of the circular base of an inverted cone of height $h$ and base radius $R$. The cone is capped by a hemisphere of radius $R$ as shown in the figure. The electric flux through the conical surface is $\frac{n q}{6 \epsilon_0}$ (in $SI$ units). The value of $n$ is. . . .

Pick out the statement which is incorrect.

$A$ 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:

$A$ point charge causes an electric flux of $-2 \times 10^4 \ Nm^2 C^{-1}$ to pass through a spherical Gaussian surface of $8.0 \ cm$ radius,centred on the charge. The value of the point charge is: (Given $\epsilon_0 = 8.85 \times 10^{-12} \ C^2 N^{-1} m^{-2}$)

$A$ few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that:
$(A)$ $|Q_1| > |Q_2|$
$(B)$ $|Q_1| < |Q_2|$
$(C)$ at a finite distance to the left of $Q_1$ the electric field is zero
$(D)$ at a finite distance to the right of $Q_2$ the electric field is zero

$A$ positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is the flux through a closed Gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is the flux through a closed Gaussian surface $S_2$ whose radius is just greater than $R_1$,then: