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

The Correct Option is D

To determine the potential difference between two points \( C \) and \( P \) on a charged thin spherical shell, we need to consider the properties of potential inside and outside the shell. For a thin spherical shell:

• Inside the shell (any point within the radius of the shell), the electric potential is constant. This is because the electric field inside the shell is zero due to the symmetry and Gauss's law. Therefore, the potential difference between any two points inside the shell is zero.
• Outside the shell, the shell behaves as a point charge located at its center. The potential \( V \) at a distance \( r \) from the center is given by \( V = \frac{kQ}{r} \), where \( Q \) is the total charge on the shell, and \( k = 9 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
• If the points \( C \) and \( P \) are both either inside or at the same radius from the center of the shell, the potential difference between them is zero.

Given that both points are either inside the shell or at the same radial distance, the potential difference is: Zero

Answer 2

Step 1: Recall the Property of a Spherical Shell 

Inside a charged spherical shell, the electric field is zero due to the symmetry of the charge distribution. Hence, the potential remains constant inside the shell.

Step 2: Potential Difference Between C and P

Since the electric field is zero inside the shell, the potential at any two points inside the shell, including C and P, is the same:

$$ V_C = V_P, \quad \Delta V = V_C - V_P = 0 $$

Step 3: Conclude

The potential difference between C and P is 0V.