1. For one sphere in air

Since the two identical spheres are symmetrically suspended, for each sphere:

• Weight acts downward: $mg$
• Electrostatic repulsion acts horizontally: $F_e$
• Tension acts along the string

If each string makes angle $\alpha$ with the vertical, then the angle between strings is $2\alpha = \theta$

From equilibrium of one sphere, $T\sin\alpha = F_e, \qquad T\cos\alpha = mg$ So, $\tan\alpha = \frac{F_e}{mg}$

2. For one sphere in water

When immersed in water:

• Electrostatic force reduces by dielectric constant $K$: $F_e' = \frac{F_e}{K}$
• Apparent weight decreases due to buoyancy.

If density of sphere material is $\rho_s = 1.5\,\text{g/cc}$ and density of water is $\rho_w = 1\,\text{g/cc}$, then $mg = \rho_s V g$ Buoyant force: $B = \rho_w V g$ Hence apparent weight in water is $W' = mg - B = (\rho_s - \rho_w)Vg$

So, $W' = (1.5-1)Vg = 0.5Vg$ while in air, $mg = 1.5Vg$

Thus, $\frac{W'}{mg} = \frac{0.5}{1.5} = \frac13$ So, $W' = \frac{mg}{3}$

3. Given angle remains same

Since the angle remains the same, $\alpha$ remains the same. Therefore, $\tan\alpha = \frac{F_e}{mg} = \frac{F_e'}{W'}$ Substitute $F_e' = \dfrac{F_e}{K}$ and $W' = \dfrac{mg}{3}$: $\frac{F_e}{mg} = \frac{F_e/K}{mg/3}$ $\frac{F_e}{mg} = \frac{3F_e}{Kmg}$ Cancelling $F_e$ and $mg$, $1 = \frac{3}{K}$ $K = 3$

4. Final answer

The dielectric constant of water is $\boxed{3}$