Two metal spheres $A$ and $B$ of radii $a\&b(a<b)$ respectively are at a large distance apart. Each sphere carries a charge of $100\mu \mathrm{C}.$ The spheres are connected by a conducting wire, then

The charge $\mathrm{q}$ is fired towards another charged particle $\mathrm{Q}$ which is fixed, with a speed $\mathrm{v}$. It approaches $\mathrm{Q}$ up to closest distance $\mathrm{r}$ and then returns. If $\mathrm{q}$ were given a speed $2\mathrm{v}$, the closest distance of approach would be

$\mathrm{ABC}$ is a right angled triangle situated in a uniform electric field $\overrightarrow{\mathrm{E}}$ which is in the plane of the triangle. The points $\mathrm{A}$ and $\mathrm{B}$ are at the same potential of $15 \mathrm{V}$ while the point $\mathrm{C}$ is at a potential of $20 \mathrm{V}.$ $\mathrm{AB}=3 \mathrm{cm}$ and $\mathrm{BC}=4 \mathrm{cm}.$ The magnitude of electric field is (in S.I. Units)

The work which is required to be done to make an arrangement of four particles each having a charge $+q$ such that the particles lie at the four corners of a square of side $a$ is

Three identical charged capacitors each of capacitance $5 \mathrm{\mu F}$ are connected as shown in figure. Potential difference across capacitor $\left(3\right),$ long time after the switches ${\mathrm{K}}_1$ and ${\mathrm{K}}_2$ are closed, is