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:

$(\text{Take}\frac{1}{4\mathrm{\pi <!--\; \pi \; -->}{\mathrm{\epsilon <!--\; \epsilon \; -->}}_0}=9\times <!--\; \times \; -->{10}^{9}SIunits)$

The correct answer is Zero.

Step-by-Step Explanation:

1. Analyzing the Given Information:
From the provided image, we can identify the following parameters of the system:
- A thin spherical shell of radius $R=3\text{cm}$ is charged with a total charge $q=1\text{μC}$.
- Point C represents the center of the spherical shell.
- Point P represents a point located on the surface of the spherical shell.

2. Electrostatic Properties of a Thin Spherical Shell:
According to Gauss's Law, the electric field inside a hollow charged spherical shell is zero at all interior points:

$E_{\text{inside}}=0$

Since the electric field $E$ and electric potential $V$ are related by the gradient relationship:

$E=-\frac{dV}{dr}$

A zero electric field means that the rate of change of potential with respect to distance is zero. Consequently, the electric potential is constant everywhere inside the shell and is equal to the potential at the surface:

$V=\text{constant for all}r\le R$

3. Finding the Potential at C and P:
The potential at the center C of the shell ($r=0$) is:

$V_C=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$

The potential at the surface point P ($r=R$) is:

$V_P=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}$

4. Calculating the Potential Difference:
The potential difference between points C and P is given by:

$V_C-V_P=\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}-\frac{1}{4\pi {\epsilon }_0}\frac{q}{R}=0$

Therefore, the potential difference is Zero.