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Step-by-Step Solution
Step 1: Understand the Physical Scenario
Two identical charged spheres are suspended by threads of equal length, making an angle \theta with each other in air. When the entire arrangement is moved into water, the angle \theta remains the same, despite changes in forces due to buoyancy and the dielectric properties of water. We need to find the dielectric constant of water under these conditions.
Step 2: Forces in Air
Weight ( W ): Each sphere has a weight mg .
Electrostatic force ( F_{\text{air}} ): In air, the repulsive force between each pair of similarly charged spheres is
F_{\text{air}} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}.
Tension in the threads: Balances both the horizontal (electrostatic) and vertical (weight) components when the system is at equilibrium, setting the angle \theta .
The key ratio that determines the angle \theta is the ratio of the electrostatic force to the weight, \frac{F_{\text{air}}}{mg} .
Step 3: Forces in Water
Effective weight ( W_{\text{effective}} ): Due to buoyancy, the effective weight of each sphere in water is reduced. If
\rho_{\text{sphere}} is the density of the spheres, \rho_{\text{water}} is the density of water, and V is the volume of each sphere, then
W_{\text{effective}} = mg - \rho_{\text{water}} V g
= (\rho_{\text{sphere}} - \rho_{\text{water}}) V g.
Electrostatic force in water ( F_{\text{water}} ): Water has dielectric constant K , so
F_{\text{water}} = \frac{F_{\text{air}}}{K}.
For the angle to remain the same, the ratio of electrostatic force to the effective weight in water must remain equal to the corresponding ratio in air. Thus,
\frac{F_{\text{air}}}{mg} = \frac{F_{\text{water}}}{(\rho_{\text{sphere}} - \rho_{\text{water}}) V g}.
Step 4: Equating the Ratios
Substitute F_{\text{water}} = \frac{F_{\text{air}}}{K} and m = \rho_{\text{sphere}} V into the above relation. We get:
\frac{F_{\text{air}}}{\rho_{\text{sphere}} V g}
=
\frac{\frac{F_{\text{air}}}{K}}{(\rho_{\text{sphere}} - \rho_{\text{water}}) V g}.
Cancel F_{\text{air}} and V g on both sides:
\frac{1}{\rho_{\text{sphere}}} = \frac{1/K}{(\rho_{\text{sphere}} - \rho_{\text{water}})}.
Rewriting for K :
K = \frac{\rho_{\text{sphere}}}{\rho_{\text{sphere}} - \rho_{\text{water}}}.
Step 5: Numerical Substitution and Final Answer
Given densities: \rho_{\text{sphere}} = 1.5 \, \text{g/cc} and \rho_{\text{water}} = 1.0 \, \text{g/cc}. Substitute these values into the formula for K :
K = \frac{1.5}{1.5 - 1.0} = \frac{1.5}{0.5} = 3.
Hence, the dielectric constant of water in this scenario is 3.
