Question

Two sphere are connected with a conducting wire. $E_{S_1}$ & $E_{S_2}$ are electric field at surface of sphere at equilibrium, find $\frac{E_{S_1}}{E_{S_2}}$.

• A. 4.5
• B. 1.125
• C. 2.25
• D. 7.5

Answer 1

The Correct Option is C

Step 1: Understand the equilibrium condition for connected conductors.
When connected by a conducting wire, the potentials of the two spheres become equal: $V_1 = V_2$.

Step 2: Relate potentials to charges and radii.
$\frac{k q_1}{r_1} = \frac{k q_2}{r_2} \implies \frac{q_1}{q_2} = \frac{r_1}{r_2}$.

Step 3: Write the expression for the electric field at the surface.
$E = \frac{k q}{r^2}$.

Step 4: Find the ratio of electric fields.
$\frac{E_{S_1}}{E_{S_2}} = \frac{k q_1 / r_1^2}{k q_2 / r_2^2} = \left( \frac{q_1}{q_2} \right) \left( \frac{r_2}{r_1} \right)^2$.

Step 5: Substitute the charge ratio into the electric field ratio.
$\frac{E_{S_1}}{E_{S_2}} = \left( \frac{r_1}{r_2} \right) \left( \frac{r_2}{r_1} \right)^2 = \frac{r_2}{r_1}$.

Step 6: Plug in the values $r_1 = 8 \text{ cm}$ and $r_2 = 18 \text{ cm}$.
$\frac{E_{S_1}}{E_{S_2}} = \frac{18}{8} = \frac{9}{4} = 2.25$.

Final Answer: The ratio is 2.25. Correct option is (3).