If $\epsilon_0$ denotes the permittivity of free space and $\Phi_E$ is the flux of the electric field through the area bounded by the closed surface, then dimensions of $\left(\epsilon_0 \frac{d \phi_E}{d t}\right)$ are that of :

Two charges $q_1$ and $q_2$ are separated by a distance of 30 cm . A third charge $q_3$ initially at ' C ' as shown in the figure, is moved along the circular path of radius 40 cm from C to D . If the difference in potential energy due to movement of $q_3$ from C to D is given by $\frac{q_3 \mathrm{~K}}{4 \pi \epsilon_0}$, the value of K is :

A metallic ring is uniformly charged as shown in figure. AC and BD are two mutually perpendicular diameters. Electric field due to arc $A B$ at ' $O$ ' is ' $E$ ' in magnitude. What would be the magnitude of electric field at ' O ' due to arc ABC ?

Two infinite identical charged sheets and a charged spherical body of charge density ' $\rho$ ' are arranged as shown in figure. Then the correct relation between the electrical fields at $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and D points is: