Question

Two hollow conducting spheres of radii R₁ and R₂ (R₁ >>R₂) have equal charges. The potential would be:

• A. More on bigger sphere
• B. More on smaller sphere
• C. Equal on both the sphere
• D. Dependent on the material properly of the sphere

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a hollow conducting sphere, the electric potential (\(V\)) on its surface is uniform and is determined by the total charge and the radius of the sphere.
Key Formula or Approach:
The potential \(V\) at the surface of a sphere of radius \(R\) carrying charge \(Q\) is given by:
\[ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{R} \]
Step 2: Detailed Explanation:
Given that both spheres have equal charges (\(Q_1 = Q_2 = Q\)).
The potential of the first sphere is \(V_1 = \frac{1}{4\pi\epsilon_0} \frac{Q}{R_1}\).
The potential of the second sphere is \(V_2 = \frac{1}{4\pi\epsilon_0} \frac{Q}{R_2}\).
From these equations, it is clear that for a constant charge \(Q\):
\[ V \propto \frac{1}{R} \]
Since it is given that \(R_1 >> R_2\), it follows that:
\[ \frac{1}{R_2}>\frac{1}{R_1} \implies V_2>V_1 \]
Thus, the smaller sphere (\(R_2\)) has a higher potential.
Step 3: Final Answer:
The potential will be more on the smaller sphere.