$A$ ball of mass $1 \, g$ and charge $10^{-8} \, C$ moves from point $A$ $(V_A = 600 \, V)$ to point $B$ where the potential is zero. The velocity of the ball at point $B$ is $20 \, cm/s$. Find the velocity of the ball at point $A$ in $cm/s$.

$A$ charge $q$ is fired towards another charged particle $Q$,which is fixed,with a speed $v$. It approaches $Q$ up to a closest distance $r$ and then returns. If $q$ were given a speed $2v$,the closest distance of approach would be

Two point charges of $12\ \mu C$ and $8\ \mu C$ are placed $10\ cm$ apart. The work done in bringing them $4\ cm$ closer is:

$A$ particle of mass $100\, g$ and charge $2\, \mu C$ is released from a distance of $50\, cm$ from a fixed charge of $5\, \mu C$. Find the speed of the particle when its distance from the fixed charge becomes $3\, m$. Neglect any other force. (Result in $m/s$) (in $.73$)

$A$ ball of mass $1\, g$ and charge $10^{-8}\ C$ moves from point $A$ at a potential of $600\, V$ to point $B$ at a potential of $0\, V$. If the velocity of the ball at point $B$ is $20\, cm/s$,then the velocity of the ball at point $A$ is . . . . . . $cm/s$.

The distance between the plates of a capacitor is $1 \ mm$. An electric field of $5 \times 10^5 \ V/m$ is produced between the two plates. What will be the change in the potential energy $(PE)$ of an electron if it is moved from one plate to the other?