The electric potential due to a point charge q at a distance r is given by V = kq/r. When the charge is doubled to 2q and distributed uniformly on a hollow sphere of radius 4r, the potential at any point outside the sphere is the same as that of a point charge at its center. Therefore, the new potential at point P (distance r from center) is V' = k(2q)/r = 2(kq/r) = 2V. The problem states this new potential is V/n, so V/n = 2V, meaning n = 2.

Point Charge Potential: $V=\frac{kq}{r}$

Sphere Potential (outside): $V^{\prime }=\frac{k\left(2q\right)}{r}=2V$

Final Answer: 2