Question

Two metallic spheres of radii in the ratio 1 : 2 are charged and joined by a connecting wire. The ratio of electric field intensities at the surfaces of the spheres is

• A. 2 : 1
• B. 1 : 2
• C. 1 : 3
• D. 3 : 1
• E. 2 : 3

Hint:

For connected conductors, the surface electric field is inversely proportional to the radius of curvature ($E \propto 1/R$). Smaller spheres have stronger surface fields.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When two conductive spheres are joined by a connecting wire, charge flows until they reach electrostatic equilibrium. At this point, their electric potentials become equal (\(V_1 = V_2\)).
Step 2: Key Formula or Approach:
The potential at the surface of a sphere is \(V = \frac{kQ}{R}\).
The electric field at the surface is \(E = \frac{kQ}{R^2} = \frac{V}{R}\).
Since \(V\) is identical for both, electric field is inversely proportional to the radius: \(E \propto \frac{1}{R}\).
Step 3: Detailed Explanation:
Since the spheres are connected, \(V_1 = V_2 = V\).
The electric field for sphere 1 is \(E_1 = \frac{V}{R_1}\).
The electric field for sphere 2 is \(E_2 = \frac{V}{R_2}\).
Find the ratio of their electric fields:
\[ \frac{E_1}{E_2} = \frac{V / R_1}{V / R_2} = \frac{R_2}{R_1} \] We are given that the ratio of the radii is \(\frac{R_1}{R_2} = \frac{1}{2}\).
Therefore, inverting this ratio gives:
\[ \frac{E_1}{E_2} = \frac{2}{1} \] The ratio of electric fields is 2 : 1.
Step 4: Final Answer:
The ratio is 2 : 1.