Why gravitational forces or spring forces are conservative forces ?
When an external force does work in taking a body from a point to another against a force like spring force or gravitational force, the work gets stored as potential energy of the body and when the external force is removed the body moves gaining kinetic energy and losing an equal amount of potential energy. The sum of kinetic energy and potential energies remains constant. Forces of these kinds are called conservative forces.
For examples : Spring force, gravitational force, electrostatic force, magnetic force etc.
At a distance $l$ from a uniformly charged long wire, a charged particle is thrown radially outward with a velocity $u$ in the direction perpendicular to the wire. When the particle reaches a distance $2 l$ from the wire, its speed is found to be $\sqrt{2} u$. The magnitude of the velocity, when it is a distance $4 l$ away from the wire is (ignore gravity)
A simple pendulum with a bob of mass $m = 1\ kg$ , charge $q = 5\mu C$ and string length $l = 1\ m$ is given a horizontal velocity $u$ in a uniform electric field $E = 2 × 10^6\ V/m$ at its bottom most point $A$ , as shown in figure. It is given a speed $u$ such that the particle leave the circular path at its topmost point $C$ . Find the speed $u$ . (Take $g = 10\ m/s^2$ )
A particle of mass $m$ having negative charge $q$ move along an ellipse around a fixed positive charge $Q$ so that its maximum and minimum distances from fixed charge are equal to $r_1$ and $r_2$ respectively. The angular momentum $L$ of this particle is
Figure shows a charge array known as an electric quadrupole. For a point on the axis of the quadrupole, obtain the dependence of potential on $r$ for $r / a>>1,$ and contrast your results with that due to an electric dipole, and an electric monopole (i.e., a single charge).
Charge $q_{2}$ is at the centre of a circular path with radius $r$. Work done in carrying charge $q_{1}$, once around this equipotential path, would be