An electron revolves around an infinite cylindrical wire having uniform linear change density $2 \times 10^{-8}\,Cm ^{-1}$ in circular path under the influence of attractive electrostatic field as shown in the figure. The velocity of electron with which it is revolving is $.........\times 10^6\,ms ^{-1}$. Given mass of electron $=9 \times 10^{-31}\,kg$

As shown in figure, a cuboid lies in a region with electric field $E=2 x^2 \hat{i}-4 y \hat{j}+6 \hat{k} \quad N / C$. The magnitude of charge within the cuboid is $n \varepsilon_0 C$. The value of $n$ is $............$ (if dimension of cuboid is $1 \times 2 \times 3 \;m ^3$ )

A circular disc of radius $R$ carries surface charge density $\sigma(r)=\sigma_0\left(1-\frac{r}{R}\right)$, where $\sigma_0$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $\phi_0$. Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $\phi$. Then the ratio $\frac{\phi_0}{\phi}$ is. . . . . .

Each of two large conducting parallel plates has one sided surface area $A$. If one of the plates is given a charge $Q$ whereas the other is neutral, then the electric field at a point in between the plates is given by

A charge $q$ is placed at the centre of the open end of cylindrical vessel. The flux of the electric field through the surface of the vessel is

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 $\Phi=\frac{ Q }{5 \epsilon_0}$