$A$ particle of mass $2\,g$ and charge $1\,\mu C$ is released from a distance of $1\,m$ from a fixed charge of $1\,mC$. What will be the velocity of the particle at a distance of $10\,m$ from the fixed charge (in $m/s$)?

In the given square $ABCD$,charges $q$,$2q$,$3q$,and $4q$ are placed at corners $A$,$B$,$C$,and $D$ respectively. What is the direction of the net electric field at the center $O$?

Consider the points lying on a straight line joining two fixed opposite charges. Between the charges,there is:

Two equal charges $q$ are placed at a distance of $2a$ on the $x$-axis. A third charge $q$ of mass $m$ is placed at the midpoint between them. If this charge is displaced by a small distance $x$ from the equilibrium position, the particle will:

Two identical conducting spheres $A$ and $B$ carry equal charge. They are separated by a distance much larger than their diameter,and the force between them is $F$. $A$ third identical conducting sphere,$C$,is uncharged. Sphere $C$ is first touched to $A$,then to $B$,and then removed. As a result,the force between $A$ and $B$ would be equal to

The adjacent diagram shows a charge $+Q$ held on an insulating support $S$ and enclosed by a hollow spherical conductor. $O$ represents the centre of the spherical conductor,and $P$ is a point such that $OP = x$ and $SP = r$. The electric field at point $P$ will be