Two electrons are moving towards each other, each with a velocity of $10^6 \,m / s$. What will be closest distance of approach between them is ......... $m$
$1.53 \times 10^{-8}$
$2.53 \times 10^{-10}$
$2.53 \times 10^{-6}$
$0$
Consider the configuration of a system of four charges each of value $+q$ . The work done by external agent in changing the configuration of the system from figure $(1)$ to figure $(2)$ is
The work done in carrying a charge of $5\,\mu \,C$ from a point $A$ to a point $B$ in an electric field is $10\,mJ$. The potential difference $({V_B} - {V_A})$ is then
A pellet carrying charge of $0.5\, coulombs$ is accelerated through a potential of $2,000\, volts$. It attains a kinetic energy equal to
A problem of practical interest is to make a beam of electrons turn at $90^o$ corner. This can be done with the electric field present between the parallel plates as shown in the figure. An electron with kinetic energy $8.0 × 10^{-17}\ J$ enters through a small hole in the bottom plate. The strength of electric field that is needed if the electron is to emerge from an exit hole $1.0\ cm$ away from the entrance hole, traveling at right angles to its original direction is $y × 10^5\ N/C$ . The value of $y$ is
Kinetic energy of an electron accelerated in a potential difference of $100\, V$ is