How much kinetic energy will be gained by an $\alpha - $particle in going from a point at $70\,V$ to another point at $50\,V$

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 $g = 10\ m/s^2$ )

Two identical thin rings, each of radius $R $ meter are coaxially placed at distance $R$ meter apart. If $Q_1$ and $Q_2$ coulomb are respectively the charges uniformly spread on the two rings, the work done in moving a charge $q$ from the centre of one ring to that of the other is

A test charge $q$ is made to move in the electric field of a point charge $Q$ along two different closed paths as per figure. First path has sections along and perpendicular to lines of electric field. Second path is a rectangular loop of the same area as the first loop. How does the work done compare in the two cases ?

If $OP = 1\,\,cm$ and $OS = 2\,\, cm$, work done by electric field in shifting a point charge $\frac {4\sqrt 2}{27}\,\, μC$ from point $P$ to $S$ in given figure is

The displacement of a charge $Q$ in the electric field $E = {e_1}\hat i + {e_2}\hat j + {e_3}\hat k$ is $\hat r = a\hat i + b\hat j$. The work done is