Question

Two charged conducting spheres \(S_1\) and \(S_2\) of radii \(8\) cm and \(18\) cm are connected to each other by a wire. After equilibrium is established, the ratio of electric fields on \(S_1\) and \(S_2\) spheres are \(E_{S_1}\) and \(E_{S_2}\) respectively. The value of \( \dfrac{E_{S_1}}{E_{S_2}} \) is:

• A. \( \dfrac{3}{2} \)
• B. \( \dfrac{2}{3} \)
• C. \( \dfrac{4}{9} \)
• D. \( \dfrac{9}{4} \)

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
When two conducting spheres are connected by a wire, charge flows between them until they reach the same electrical potential. At equilibrium, \(V_1 = V_2\).
Step 2: Key Formula or Approach:
1. Potential of a sphere: \(V = \frac{kq}{r}\)
2. Electric field at surface: \(E = \frac{kq}{r^2}\)
Step 3: Detailed Explanation:
At equilibrium, potentials are equal:
\[ V_1 = V_2 \implies \frac{kq_1}{r_1} = \frac{kq_2}{r_2} \implies \frac{q_1}{r_1} = \frac{q_2}{r_2} \]
Now, consider the ratio of electric fields:
\[ \frac{E_{S1}}{E_{S2}} = \frac{kq_1/r_1^2}{kq_2/r_2^2} = \left( \frac{q_1}{q_2} \right) \left( \frac{r_2}{r_1} \right)^2 \]
From the potential equality, substitute \(\frac{q_1}{q_2} = \frac{r_1}{r_2}\):
\[ \frac{E_{S1}}{E_{S2}} = \left( \frac{r_1}{r_2} \right) \left( \frac{r_2}{r_1} \right)^2 = \frac{r_2}{r_1} \]
Given radii \(r_1 = 8 \text{ cm}\) and \(r_2 = 18 \text{ cm}\):
\[ \frac{E_{S1}}{E_{S2}} = \frac{18}{8} = \frac{9}{4} \]
Step 4: Final Answer:
The ratio of the electric fields is 9 / 4.