$A$ proton is accelerated through $50,000 \, V$. Its energy will increase by

If the electric potential of the inner metal sphere of radius $a$ is $10 \text{ V}$ and that of the outer spherical shell of radius $b$ is $5 \text{ V}$,then the potential at the centre will be ...... $\text{V}$.

An electron (charge = $1.6 \times 10^{-19} \text{ C}$) is accelerated through a potential difference of $1,00,000 \text{ V}$. The energy acquired by the electron is:

Two electric charges $12\,\mu C$ and $-6\,\mu C$ are placed $20\, cm$ apart in air. There will be a point $P$ on the line joining these charges and outside the region between them,at which the electric potential is zero. The distance of $P$ from the $-6\,\mu C$ charge is.......$m$.

Energy of electrons can be increased by allowing them

Along the $x$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at a point $P$ located at a distance $r$ from the charge $-q$ $(r > a)$ is ($\varepsilon_0$ is the permittivity of free space):