Charges $-q,\, q,\,q$ are placed at the vertices $A$, $B$, $C$ respectively of an equilateral triangle of side $'a'$ as shown in the figure. If charge $-q$ is released keeping remaining two charges fixed, then the kinetic energy of charge $(-q)$ at the instant when it passes through the mid point $M$ of side $BC$ is
$\frac{{{q^2}}}{{8\pi { \in _0}a}}$
$\frac{{{q^2}}}{{4\pi { \in _0}a}}$
$\frac{{{q^2}}}{{2\pi { \in _0}a}}$
$\frac{{{q^2}}}{{\pi { \in _0}a}}$
There is an electric field $E$ in $X$-direction. If the work done on moving a charge $0.2\,C$ through a distance of $2\,m$ along a line making an angle $60^\circ $ with the $X$-axis is $4.0\;J$, what is the value of $E$........ $N/C$
State which of the following is correct
A point charge $q$ of mass $m$ is suspended vertically by a string of length $l$. A point dipole of dipole moment $\overrightarrow{ p }$ is now brought towards $q$ from infinity so that the charge moves away. The final equilibrium position of the system including the direction of the dipole, the angles and distances is shown in the figure below. If the work done in bringing the dipole to this position is $N \times( mgh )$, where $g$ is the acceleration due to gravity, then the value of $N$ is. . . . . . (Note that for three coplanar forces keeping a point mass in equilibrium, $\frac{F}{\sin \theta}$ is the same for all forces, where $F$ is any one of the forces and $\theta$ is the angle between the other two forces)
A ball of mass $1\, g$ and charge ${10^{ - 8}}\,C$ moves from a point $A$. where potential is $600\, volt$ to the point $B$ where potential is zero. Velocity of the ball at the point $B$ is $20\, cm/s$. The velocity of the ball at the point $A$ will be
Explain electric potential energy. Show that the sum of kinetic energy and electric potential energy remains constant.