A uniform electric field of $10\,N / C$ is created between two parallel charged plates (as shown in figure). An electron enters the field symmetrically between the plates with a kinetic energy $0.5\,eV$. The length of each plate is $10\,cm$. The angle $(\theta)$ of deviation of the path of electron as it comes out of the field is $.........$(in degree).
$44$
$43$
$42$
$45$
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)$
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
There is a uniform electric field of strength ${10^3}\,V/m$ along $y$-axis. A body of mass $1\,g$ and charge $10^{-6}\,C$ is projected into the field from origin along the positive $x$-axis with a velocity $10\,m/s$. Its speed in $m/s$ after $10\,s$ is (Neglect gravitation)
An electron falls through a small distance in a uniform electric field of magnitude $2 \times {10^4}N{C^{ - 1}}$. The direction of the field is reversed keeping the magnitude unchanged and a proton falls through the same distance. The time of fall will be
A simple pendulum is suspended in a lift which is going up with an acceleration $5\ m/s^2$. An electric field of magnitude $5 \ N/C$ and directed vertically upward is also present in the lift. The charge of the bob is $1\ mC$ and mass is $1\ mg$. Taking $g = \pi^2$ and length of the simple pendulum $1\ m$, the time period of the simple pendulum is ......$s$