Continuous Charge Distribution (NCERT) - Free MCQ Practice Test with solutions,

Linear charge density, λ = charge/length ; A → q
Surface charge density, σ = charge/area; B → r
Volume charge density, ρ = charge/volume; C → p

Here D = 2r = 4.4 m,
or r = 2.2 m, σ = 60 μC m^-2
Charge on the sphere, q = σ x 4πr²
= 60 x 10^-6 x 4 x (22/7) x (2.2)² = 3.7 x 10^-3 C.

When a charge +q is placed at the centre of spherical cavity as shown in figure.
Charge induced on the inner surface of shell = - q ... (i)
Charge induced on the outer surface of shell = + q ... (ii)

∴ Surface charge dcnsity on the inner surface = -q/4πR²₁

The outer surface of the outer shell similarly has its charge uniformly distributed. Thus, the surface charge density, σ₂, is given by σ₂ = q₂ / (4πR₂²).

At the centre of the ring, E = 0 when a positive charge q > 0 is displaced away from the centre in the plane of the ring, say to the right, force of repulsion on q, due to charge on right half increases and due to charge on left half decreases. Therefore, charge q is pushed back towards the centre. So option (a) is correct.
When charge q is negative (q < 0), force is of attraction.
Therefore, charge q displaced to the right continues moving to the right till it hits the ring. Along the axis of the ring, at a distance r from the centre.
E = (Qr/(4πε₀(r² +a²)^3/2)
If charge q is negative (q < 0), it will perform SHM for small displacement along the axis.