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 constant. Find the potential at the centre $O$ of ring. [in volt]

A neutral spherical copper particle has a radius of $10 \,nm \left(1 \,nm =10^{-9} \,m \right)$. It gets charged by applying the voltage slowly adding one electron at a time. Then, the graph of the total charge on the particle versus the applied voltage would look like

The potential at a distance $R/2$ from the centre of a conducting sphere of radius $ R$ will be

If the potential of the inner shell is $10\,V$ and that of the outer shell is $5\,V$, then potential at the centre will be....$V$

Consider a sphere of radius $R$ with uniform charge density and total charge $Q$. The electrostatic potential distribution inside the sphere is given by $\theta_{(r)}=\frac{Q}{4 \pi \varepsilon_{0} R}\left(a+b(r / R)^{C}\right)$. Note that the zero of potential is at infinity. The values of $(a, b, c)$ are

Two equal positive point charges are kept at points $A$ and $B$ . The electric potential, while moving from $A$ to $B$ along straight line