$A$ proton is accelerated through a potential difference of $1 \,V$. The kinetic energy $(KE)$ of the proton will be $....... \,eV$.

At the centre of a half ring of radius $R=10 \ cm$ and linear charge density $\lambda = 4 \ nC \ m^{-1}$,the potential is $x \pi \ V$. The value of $x$ is . . . . .

$A$ spherical drop of liquid carrying charge $Q$ has potential $V_0$ at its surface. If two drops of same charge and radius combine to form a single spherical drop,then the potential at the surface of the new drop is (Assume $V=0$ at infinity.)

Two charges $3 \times 10^{-8} \; C$ and $-2 \times 10^{-8} \; C$ are located $15 \; cm$ apart. At what point on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.

The linear charge density on a dielectric ring of radius $R$ varies with $\theta$ as $\lambda = \lambda_0 \cos(\theta/2)$,where $\lambda_0$ is a constant. Find the potential at the centre $O$ of the ring.

Concentric metallic hollow spheres of radii $R$ and $4R$ hold charges $Q_1$ and $Q_2$ respectively. Given that surface charge densities of the concentric spheres are equal,the potential difference $V(R) - V(4R)$ is