Question: 35. A thin spherical shell is charged by some source. The potential difference between the two points C and P (in V) shown in the figure is :
(Take \frac{1}{4 \pi \epsilon_{0}}=9 \times 10^{9} SI units)
(1) 3 \times 10^{5}
(2) 1 \times 10^{5}
(3) 0.5 \times 10^{5}
(4) zero
Answer: Option (4)
Explanation:
The given object is a thin spherical shell uniformly charged with total charge q.
Point C is the centre of the spherical shell and point P lies on the surface of the shell.
For a charged spherical shell, the electric potential at any point inside the shell is constant and equal to the potential at the surface.
The potential at the surface of a charged spherical shell is given by
V = \frac{1}{4\pi\epsilon_0}\frac{q}{R}The potential at the centre C is also
V_C = \frac{1}{4\pi\epsilon_0}\frac{q}{R}The potential at point P on the surface is
V_P = \frac{1}{4\pi\epsilon_0}\frac{q}{R}Thus,
V_C = V_PThe potential difference between points C and P is
\Delta V = V_C - V_P = 0Hence, the correct answer is Option (4).