A Charged particle of mass $m$ and charge $q$ is released from rest in a uniform electric field $E.$ Neglecting the effect of gravity, the kinetic energy of the charged particle after $'t'$ second is
$\frac{{E{q^2}m}}{{2t}}$
$\frac{{2{E^2}{t^2}}}{{mq}}$
$\frac{{{E^2}{q^2}{t^2}}}{{2m}}$
$\frac{{Eqm}}{t}$
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)$
An electric line of force in $X$, $Y-$ plane is given by $x^2+y^2 = 1$. A particle with unit positive charge, initially at rest at the point $x = 1, y = 0$ in the $X, Y-$ plane
An electron is moving towards $x$-axis. An electric field is along $y$-direction then path of electron is
A particle of charge $1\ \mu C\ \&\ mass$ $1\ gm$ moving with a velocity of $4\ m/s$ is subjected to a uniform electric field of magnitude $300\ V/m$ for $10\ sec$. Then it's final speed cannot be.......$m/s$
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