$125$ identical charged small spheres coalesce to form a big charged sphere. If the electric potential on each small sphere is $60 \text{ mV}$, then the electric potential on the bigger sphere formed is: (in $\text{ V}$)

Two equal positive charges each of value $q$ are placed at points $A$ and $B$, where $AB = 3x$. A third charge $-3q$ is placed at point $C$ at a distance $x$ from $A$ on the line segment $AB$. The potential energy of the system is nearly ($\epsilon_0$ = permittivity of free space).

Two point charges of $1 \text{ nC}$ and $2 \text{ nC}$ are placed at two corners of an equilateral triangle of side $3 \text{ cm}$. The work done in bringing a charge of $3 \text{ nC}$ from infinity to the third corner of the triangle is . . . . . . $\mu\text{J}$. $( \frac{1}{4\pi\epsilon_{0}} = 9 \times 10^{9} \text{ N.m}^{2}/\text{C}^{2} )$

Calculate the electric potential on the axis of a ring of radius $R$ due to a charge $Q$ uniformly distributed along it.

$A$ $200 \text{ J}$ of work is done in moving a charge of $5 \text{ C}$ from a point $A$ where the potential is $-20 \text{ V}$ to another point $B$ where the potential is $V \text{ V}$. The value of $V$ at point $B$ is: (in $text{ V}$)

$A$ uniformly charged solid sphere of radius $R$ has potential $V_0$ (measured with respect to $\infty$) on its surface. For this sphere, the equipotential surfaces with potentials $\frac{3V_0}{2}, \frac{5V_0}{4}, \frac{3V_0}{4}$, and $\frac{V_0}{4}$ have radii $R_1, R_2, R_3$, and $R_4$ respectively. Then: