Question

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 × 109$$\frac{1}{4\pi\epsilon_0}=9\times10^9 SI units)

• A. 3 × 105
• B. 1 × 105
• C. 0.5 × 105
• D. Zero

Answer 1

Answer: (D)

Zero

Answer 2

A charged spherical shell has the same potential at all points inside the shell and on its surface.
This property arises because the electric field inside the shell is zero.
- The potential $V$ at any point inside the shell (including the center C) and on the surface $P$ is given by:

$V = \frac{1}{4\pi\epsilon_0} \cdot \frac{q}{R}$,

where
$q = 1 \, \mu C = 1 \times 10^{-6} \,C$,
$R = 3 \, cm = 0.03 \, m$, and
$\frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \, SI units$.

- Since the potential $V$ is the same at both points $C$ and $P$,
the potential difference $\Delta V = V_P - V_C = 0$.
Thus, the potential difference between $C$ and $P$ is Zero.