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 conducting spheres are joined by a wire, charge flows until their electric potentials become equal ($V_1 = V_2$).
Key Formula or Approach:
Potential $V = \frac{kQ}{R}$.
Electric field at surface $E = \frac{kQ}{R^2} = \frac{V}{R}$.

Step 2: Detailed Explanation:
1. Since the spheres are connected, $V_1 = V_2 = V$.
2. For the first sphere: $E_1 = \frac{V}{R_1}$.
3. For the second sphere: $E_2 = \frac{V}{R_2}$.
4. The ratio of electric field intensities is:
\[ \frac{E_1}{E_2} = \frac{V/R_1}{V/R_2} = \frac{R_2}{R_1} \]
5. Given $\frac{R_1}{R_2} = \frac{1}{2} \implies \frac{R_2}{R_1} = \frac{2}{1}$.
Thus, $E_1 : E_2 = 2 : 1$.

Step 3: Final Answer:
The ratio of electric field intensities is 2 : 1.