Two identical charged spheres are suspended by strings of equal lengths. The string make an angle of $$37°$$ with each other. When suspended in a liquid of density $$0.7$$ g cm$$^{-3}$$, the angle remains same. If density of material of the sphere is $$1.4$$ g cm$$^{-3}$$, the dielectric constant of the liquid is _____ ($$\tan 37° = \frac{3}{4}$$)

In air: $$\tan(\theta/2) = \frac{F_e}{mg}$$ where $$F_e$$ is the electrostatic force.

In liquid: The angle remains the same, so $$\tan(\theta/2)$$ is unchanged. This means:

$$\frac{F_e'}{m'g} = \frac{F_e}{mg}$$

where $$F_e' = \frac{F_e}{K}$$ (dielectric constant $$K$$) and $$m'g = mg - \text{buoyancy} = mg(1 - \frac{\rho_l}{\rho_s})$$.

$$\frac{F_e/K}{mg(1 - \rho_l/\rho_s)} = \frac{F_e}{mg}$$

$$\frac{1}{K} = 1 - \frac{\rho_l}{\rho_s} = 1 - \frac{0.7}{1.4} = 1 - 0.5 = 0.5$$

$$K = 2$$.

Therefore, the answer is $$\boxed{2}$$.