A particle of charge $q$ and mass $m$ is subjected to an electric field $E = E _{0}\left(1- ax ^{2}\right)$ in the $x-$direction, where $a$ and $E _{0}$ are constants. Initially the particle was at rest at $x=0 .$ Other than the initial position the kinetic energy of the particle becomes zero when the distance of the particle from the origin is
$\sqrt{\frac{2}{a}}$
$\sqrt{\frac{1}{a}}$
$a$
$\sqrt{\frac{3}{a}}$
If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of 1 megavolt then the ratio of their kinetic energy will be
Two identical particles of mass $m$ and charge $q$ are shot at each other from a very great distance with an initial speed $v$. The distance of closest approach of these charges is
The work done in moving an electric charge $q$ in an electric field does not depend upon
$(a)$ Determine the electrostatic potential energy of a system consisting of two charges $7 \;\mu C$ and $-2\; \mu C$ (and with no external field) placed at $(-9 \;cm , 0,0)$ and $(9\; cm , 0,0)$ respectively.
$(b)$ How much work is required to separate the two charges infinitely away from each other?
$(c)$ Suppose that the same system of charges is now placed in an external electric field $E=A\left(1 / r^{2}\right) ; A=9 \times 10^{5} \;C m ^{-2} .$ What would the electrostatic energy of the configuration be?
The charge $q$ is fired towards another charged particle $Q$ which is fixed, with a speed $v$. It approaches $Q$ upto a closest distance $r$ and then returns. If $q$ were given a speed $2 v$, the closest distance of approach would be