Two spheres are connected with a conducting wire. E_s1 and E_s2 are electric fields at the surface of spheres at equilibrium. Find E_s1 /E_s2 .

The correct answer is 2.25.

Step-by-step Explanation:

From the given image, we can identify two spherical conductors with radii:
Radius of the first sphere, $R_1=8\text{cm}$
Radius of the second sphere, $R_2=18\text{cm}$

When the two spheres are connected by a conducting wire, charge flows between them until they reach electrostatic equilibrium. At equilibrium, the electric potentials at the surface of both spheres become equal:
$V_1=V_2$

The electric potential $V$ at the surface of a sphere of radius $R$ carrying a charge $Q$ is given by:
$V=\frac{kQ}{R}$

Therefore, equating the potentials gives:
$\frac{k{Q}_1}{R_1}=\frac{k{Q}_2}{R_2}$

Simplifying this expression gives the relation between the charges on the spheres:
$\frac{Q_1}{Q_2}=\frac{R_1}{R_2}$

The electric field $E$ at the surface of a sphere is given by:
$E=\frac{kQ}{R^2}$

Taking the ratio of the electric fields at the surfaces of the two spheres, we get:
$\frac{E_{s1}}{E_{s2}}=\frac{Q_1/R_1^2}{Q_2/R_2^2}=\left(\frac{Q_1}{Q_2}\right)\cdot {\left(\frac{R_2}{R_1}\right)}^2$

Substitute the charge ratio $\frac{Q_1}{Q_2}=\frac{R_1}{R_2}$ into the equation:
$\frac{E_{s1}}{E_{s2}}=\left(\frac{R_1}{R_2}\right)\cdot {\left(\frac{R_2}{R_1}\right)}^2=\frac{R_2}{R_1}$

Now, substituting the values of $R_1$ and $R_2$:
$\frac{E_{s1}}{E_{s2}}=\frac{18}{8}=2.25$