Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle $$\theta$$ with each other. When suspended in water the angle remains the same. If density of the material of the sphere is $$1.5$$ g/cc, the dielectric constant of water will be ______. (Take density of water $$= 1$$ g/cc)

Two identical charged spheres suspended by equal-length strings make angle $$\theta$$ in air. When suspended in water, the angle remains the same. Density of sphere = $$1.5$$ g/cc, density of water = $$1$$ g/cc.

In air, each sphere is in equilibrium under the action of electrostatic repulsion, gravity, and string tension; the balance of horizontal and vertical forces yields $$\tan(\theta/2) = \frac{F_e}{mg}$$, where $$F_e$$ is the electrostatic force and $$mg$$ is the weight of the sphere.

When the spheres are suspended in water, the electrostatic force is reduced by the dielectric constant $$K$$ so that $$F_e' = F_e/K$$, and the effective weight is reduced by the buoyant force: $$W' = mg - \rho_w V g = Vg(\rho_s - \rho_w)$$, where $$\rho_s$$ and $$\rho_w$$ are the densities of the sphere and water respectively and $$V$$ is the volume of the sphere. The equilibrium in water therefore gives $$\tan(\theta/2) = \frac{F_e/K}{Vg(\rho_s - \rho_w)}$$.

Since the angle remains the same in both media, we equate the two expressions: $$\frac{F_e}{mg} = \frac{F_e/K}{Vg(\rho_s - \rho_w)}$$. Noting that $$m = \rho_s V$$, this reduces to $$\frac{1}{\rho_s} = \frac{1}{K(\rho_s - \rho_w)}$$, which leads to $$K = \frac{\rho_s}{\rho_s - \rho_w} = \frac{1.5}{1.5 - 1.0} = \frac{1.5}{0.5} = 3$$.

The dielectric constant of water is 3.