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 Q1+Q2. Simillarly, the charge on the outermost shell is Q1+Q2+Q3
It is given that the surface charge densities ρ1,ρ2 and ρ3 are equal, that is,
Q1
4πR2
=
Q1+Q2
4π(2R)2
=
Q1+Q2+Q3
4π(3R)2
⇒ Q1 =
Q1+Q2
4
=
Q1+Q2+Q3
9
That is, Q1 =
Q1
Q4
+
Q2
4
⇒ Q2 = 3Q1 Therefore, Q3 = 5Q1 Hence, the ratio of the charges given to the shells Q1:Q2:Q3 is 1 : 3 : 5.