The electric field intensity at $P$ and $Q$,in the shown arrangement,are in the ratio:

Consider a uniform spherical charge distribution of radius $R_1$ centred at the origin $O$. In this distribution,a spherical cavity of radius $R_2$,centred at $P$ with distance $OP = a = R_1 - R_2$ (see figure) is made. If the electric field inside the cavity at position $\vec{r}$ is $\vec{E}(\vec{r})$,then the correct statement$(s)$ is(are):

An electric field converges at the origin whose magnitude is given by the expression $E = 100\,r\,N/C$,where $r$ is the distance measured from the origin.

$A$ charged ball $B$ hangs from a silk thread $S$,which makes an angle $\theta$ with a large charged conducting sheet $P$,as shown in the figure. The surface charge density $\sigma$ of the sheet is proportional to

The space between two large parallel plates is filled with a material of uniform charge density $\rho$. Assume that one of the plates is kept at $x=0$. The potential at any point $x$ between these plates is given by (where $A$ and $B$ are constants):

The electric field due to an infinitely long thin straight wire with uniform linear charge density of $2.5 \times 10^{-7} \ Cm^{-1}$ at a radial distance of $x$ from the wire is $7.5 \times 10^4 \ NC^{-1}$. Then $x=$ (in $cm$)