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 ……… $m$

A particle of mass $m$ and charge $q$ is kept at the top of a fixed frictionless sphere. $A$ uniform horizontal electric field $E$ is switched on. The particle looses contact with the sphere, when the line joining the center of the sphere and the particle makes an angle $45^o$ with the vertical. The ratio $\frac{qE}{mg}$ is :-

$(a)$ Calculate the potential at a point $P$ due to a charge of $4 \times 10^{-7}\; C$ located $9 \;cm$ away.

$(b)$ Hence obtain the work done in bringing a charge of $2 \times 10^{-9} \;C$ from infinity to the point $P$. Does the answer depend on the path along which the charge is brought?

A two point charges $4 q$ and $-q$ are fixed on the $x-$axis at $x=-\frac{d}{2}$ and $x=\frac{d}{2},$ respectively. If a third point charge $'q'$ is taken from the origin to $x = d$ along the semicircle as shown in the figure, the energy of the charge will

On moving a charge of $20$ coulombs by $2 \;cm , 2 \;J$ of work is done, then the potential difference between the points is (in $volt$)