1. Interpret the figure and point location

For an annular disc with outer radius $4R$ and inner radius $3R$, point $P$ is on the axis through the center. From the standard figure for this problem, $P$ is at a distance $x=4R$ from the plane of the annulus.

We need the work required to take a unit mass from $P$ to infinity.

This equals the increase in gravitational potential energy per unit mass: $W = -V(P)$ because $V(\infty)=0$.

2. Surface mass density of the annulus

Area of annulus: $A=\pi\big((4R)^2-(3R)^2\big)=\pi(16R^2-9R^2)=7\pi R^2$

Hence surface mass density is $\sigma=\frac{M}{7\pi R^2}$

3. Potential on axis due to a disc

Potential at distance $x$ on the axis of a uniform disc of radius $a$ is $V_a(x)=-2\pi G\sigma\left(\sqrt{x^2+a^2}-x\right)$ (for $x>0$).

An annulus can be treated as:

• disc of radius $4R$
• minus disc of radius $3R$

So,