$A$ sphere '$1$' with radius $R$ has charge $q$. Sphere '$2$' with radius $3R$ is far from sphere '$1$' and is initially uncharged. If the two spheres are now connected with a thin conducting wire,then the ratio $\frac{\sigma_1}{\sigma_2}$ of the surface charge densities is

$A$ spherical metal shell $A$ of radius $R_A$ and a solid metal sphere $B$ of radius $R_B < R_A$ are kept far apart and each is given charge $+Q$. Now they are connected by a thin metal wire. Then:
$(A)$ $E_A^{\text{inside}} = 0$
$(B)$ $Q_A > Q_B$
$(C)$ $\frac{\sigma_A}{\sigma_B} = \frac{R_B}{R_A}$
$(D)$ $E_A^{\text{on surface}} < E_B^{\text{on surface}}$

Assertion $:-$ The maximum charge that can be given to a conductor of fixed volume depends on its shape.
Reason $:-$ If the electric field near the conductor is sufficient for dielectric breakdown of air,no more charge can be transferred to it.

The dimensions of an atom are of the order of an $\mathring{A}$ $(10^{-10} \ m)$. Thus,there must be large electric fields between the protons and electrons. Why,then,is the electrostatic field inside a conductor zero?

$A$ point charge is kept at the centre of a metallic insulated spherical shell. Then:

Explain the construction of a Van de Graaff generator with a diagram and describe its uses.