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

To find the ratio of electric fields \(E_{S_1}\) and \(E_{S_2}\) on the surfaces of the two spheres connected by a conducting wire, we should use the concept of charge distribution and electric field on conductors.

When two conductive spheres are connected by a wire, they reach the same potential at equilibrium. For spheres of radii \(r_1\) and \(r_2\), the electric fields at their surfaces can be expressed as:

1. Electric field due to a sphere: \(E = \frac{Q}{4\pi \varepsilon_0 r^2}\) where \(Q\) is the charge, and \(r\) is the radius of the sphere.

2. Potential on a sphere: \(V = \frac{Q}{4\pi \varepsilon_0 r}\)

Since the potentials are equal (\(V_1 = V_2\)), we have:

\(\frac{Q_1}{r_1} = \frac{Q_2}{r_2}\)

Simplifying for the charges, we find:

\(Q_1 = Q_2 \frac{r_1}{r_2}\)

Thus, the electric field ratio is given by:

\(\frac{E_{S_1}}{E_{S_2}} = \frac{Q_1/r_1^2}{Q_2/r_2^2} = \frac{(Q_2 \frac{r_1}{r_2})/r_1^2}{Q_2/r_2^2} = \frac{r_2^2}{r_1^2 \cdot r_2} = \frac{r_2}{r_1}\)

Given radii \(r_1 = 8\) cm and \(r_2 = 18\) cm:

\(\frac{E_{S_1}}{E_{S_2}} = \frac{18}{8} = 2.25\)

Thus, the correct answer is 2.25.