Inside a uniformly charged ball, at $r=r_m$, the electric field is maximum. Then (Given, $\rho ={\rho }_0\left(1-\frac{r}{R}\right)$ is volume charge density)

A dipole having charges $+q$ and $–q$ is fixed on a rough surface as shown. Another charge $Q$ of mass $m$ is placed at distance $d$ from this dipole. The minimum value of $d$ for which charge $Q$ will not move. (Given $a<<d$, co-efficient of static friction ${\mu }_s=0.5$)

Two balls of charges $q_1$ and $q_2$ initially have exactly same velocity. Both the balls are, subjected to same uniform electric field for same time. As a result, the velocity of the first ball is reduced to half of its initial value and its direction changes by $60°.$ The direction of the velocity of second ball is found to change by $90°.$

The electric field and initial velocity of the charged particle are inclined at angle

Two balls of charges $q_1$ and $q_2$ initially have exactly same velocity. Both the balls are, subjected to same uniform electric field for same time. As a result, the velocity of the first ball is reduced to half of its initial value and its direction changes by $60°.$ The direction of the velocity of second ball is found to change by $90°.$

If new velocity of second charged particle has a magnitude $\text{'}x\text{'}$ times the initial velocity, then $x$ is

There is a non-conducting rod of length $L$ and negligible mass with two small balls each of mass $m$ and electric charge $Q$ attached to its ends. The rod can rotate in the horizontal plane about a fixed vertical axis crossing it a distance $\frac{L}{4}$ from one of its ends. A uniform horizontal electric field $E$ (along $+x$ axis) is established. At first the rod is in unstable equilibrium. In what position, the rod must be set so that if displaced a little from that position, it begins simple harmonic oscillations about the axis?