Solution: Potential Ratio for Semicircular Charge Distribution
Physics Principle
This problem can be solved elegantly using an analogy to a uniformly charged spherical shell. The electric potential inside a uniformly charged spherical shell is constant everywhere.
1. The “Virtual Sphere” Analogy
Consider a sphere of radius $R$ with a uniform surface charge density $\sigma$. Imagine slicing this sphere into narrow rings (strips) perpendicular to a diameter.
A strip defined by angle $\theta$ and width $d\theta$ has a radius $R\sin\theta$ and width $R d\theta$. Its area is $dA = 2\pi (R\sin\theta) R d\theta$. The charge on this strip is $dq = \sigma dA \propto \sin\theta$.
The problem specifies a linear charge distribution $\lambda = \lambda_0 \sin\theta$ on a semicircle. This mathematical form matches the charge distribution of the “slices” of a uniformly charged sphere.
2. Potential Analysis
The potential at any point inside a uniformly charged spherical shell is constant and equal to the potential at the surface ($V = \frac{kQ_{total}}{R}$).
Since the potential due to the entire sphere is constant along the diameter, and the sphere can be built from diametrically opposite pairs of such semicircular arcs (conceptually), the potential contribution from a single distribution of the form $\lambda \propto \sin\theta$ must also be constant along the diameter connecting its endpoints.
Therefore, the potential $V_O$ at the center and the potential $V_P$ at any point on the diameter $AB$ are equal.
$$ \frac{V_O}{V_P} = 1 : 1 $$