Solution
We are dealing with the flow of steady current between two spherical electrodes immersed in a large, homogeneous conducting medium.
1. Resistance of a Single Sphere
Consider a single sphere of radius $a$ in an infinite medium of resistivity $\rho$. The current density $J$ at a distance $r$ from the center is uniform radially outward:
$$ J = \frac{I}{4\pi r^2} $$The electric field is $E = \rho J$. The potential difference between the surface of the sphere ($r=a$) and infinity is:
$$ V = \int_a^\infty E \, dr = \int_a^\infty \frac{\rho I}{4\pi r^2} \, dr = \frac{\rho I}{4\pi} \left[ -\frac{1}{r} \right]_a^\infty = \frac{\rho I}{4\pi a} $$Thus, the resistance of a single isolated sphere is:
$$ R_{\text{sphere}} = \frac{V}{I} = \frac{\rho}{4\pi a} $$2. Resistance of the System
When two spheres are separated by a large distance compared to their radii, the interaction between them is negligible. The total resistance of the medium between the two spheres is the sum of the resistances of each sphere (one acting as a source, the other as a sink):
$$ R_{\text{total}} = R_1 + R_2 $$Initially, both spheres have radius $a$. So the total resistance $R_0$ is:
$$ R_0 = \frac{\rho}{4\pi a} + \frac{\rho}{4\pi a} = \frac{\rho}{2\pi a} $$The initial current $I_0$ flowing through the battery of voltage $V$ is:
$$ I_0 = \frac{V}{R_0} = \frac{V}{\frac{\rho}{2\pi a}} = \frac{2\pi a V}{\rho} $$3. Modifying the System
One sphere is replaced by another sphere of radius $\eta a$. The new total resistance $R’$ is:
$$ R’ = R_1 + R_{\text{new}} = \frac{\rho}{4\pi a} + \frac{\rho}{4\pi (\eta a)} $$ $$ R’ = \frac{\rho}{4\pi a} \left( 1 + \frac{1}{\eta} \right) = \frac{\rho}{4\pi a} \left( \frac{\eta + 1}{\eta} \right) $$The new current $I’$ is:
$$ I’ = \frac{V}{R’} = \frac{V}{\frac{\rho}{4\pi a} \left( \frac{\eta + 1}{\eta} \right)} = \frac{4\pi a \eta V}{\rho (\eta + 1)} $$4. Comparison and Calculation
Taking the ratio of $I’$ to $I_0$:
$$ \frac{I’}{I_0} = \frac{ \frac{4\pi a \eta V}{\rho (\eta + 1)} }{ \frac{2\pi a V}{\rho} } = \frac{4\pi a \eta V}{\rho (\eta + 1)} \cdot \frac{\rho}{2\pi a V} = \frac{2\eta}{\eta + 1} $$Substituting the given values:
- Initial Current $I_0 = 3.0$ A
- Ratio $\eta = 0.5$