Electrostatics

Surface charge density is ρ = $\frac{Q}{4\pi R^2}$ where R is the radius of the sphere. For cancelling the field inside the cavity due to the shell of radius R, the shell with radius 2R induces a charge -Q in its inner surfce and hence the total charge on the outer surface is $Q_1+Q_2$. Simillarly, the charge on the outermost shell is $Q_1+Q_2+Q_3$ It is given that the surface charge densities $\rho_1,\rho_2$ and $\rho_3$ are equal, that is, $\frac{Q_1}{4\pi R^2}$ = $\frac{Q_1+Q_2}{4\pi (2R)^2}$ = $\frac{Q_1+Q_2+Q_3}{4\pi (3R)^2}$ ⇒ $Q_1$ = $\frac{Q_1+Q_2}{4}$ = $\frac{Q_1+Q_2+Q_3}{9}$ That is, $Q_1$ = $\frac{Q_1}{Q_4}+\frac{Q_2}{4}$ ⇒ $Q_2$ = $3Q_1$ Therefore, $Q_3$ = $5Q_1$ Hence, the ratio of the charges given to the shells $Q_1:Q_2:Q_3$ is 1 : 3 : 5.