Four point $+ve$ charges of same magnitude $(Q)$ are placed at four corners of a rigid square frame as shown in figure. The plane of the frame is perpendicular to $Z-$ axis. If a $ -ve$ point charge is placed at a distance $z$ away from centre along axis $(z << L )$ then
$-ve$ charge oscillates along the $z-$ axis
It moves away from the frame
It moves slowly towards the frame and stays in the plane of the frame
It passes through the frame only once
An electron moving with the speed $5 \times {10^6}$ per sec is shooted parallel to the electric field of intensity $1 \times {10^3}\,N/C$. Field is responsible for the retardation of motion of electron. Now evaluate the distance travelled by the electron before coming to rest for an instant (mass of $e = 9 \times {10^{ - 31}}\,Kg.$ charge $ = 1.6 \times {10^{ - 19}}\,C)$
A proton and an $\alpha$-particle having equal kinetic energy are projected in a uniform transverse electric field as shown in figure
A uniform vertical electric field $E$ is established in the space between two large parallel plates. A small conducting sphere of mass $m$ is suspended in the field from a string of length $L$. If the sphere is given $a + q$ charge and the lower plate is charged positvely, the period of oscillation of this pendulum is :-
The surface of a planet is found to be uniformly charged. When a particle of mass $m$ and no charge is thrown at an angle from the surface of the planet, it has a parabolic trajectory as in projectile motion with horizontal range $L$. A particle of mass $m$ and charge $q$, with the same initial conditions has a range $L / 2$. The range of particle of mass $m$ and charge $2 q$, with the same initial conditions is
A proton sits at coordinates $(x, y) = (0, 0)$, and an electron at $(d, h)$, where $d >> h$. At time $t = 0$, $a$ uniform electric field $E$ of unknown magnitude but pointing in the positive $y$ direction is turned on. Assuming that $d$ is large enough that the proton-electron interaction is negligible, the $y$ coordinates of the two particles will be equal (at equal time)