JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Electrostatics

Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle

θ

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

)

Answer: 3

Solution

In air, for two identical charged spheres suspended with strings making angle

θ

tan \frac{θ}{2} = \frac{F}{m g} = \frac{q ^{2}}{4 π ε _{0} r ^{2} m g}

When submerged in water (dielectric constant

K

), the electrostatic force is reduced by a factor

K

, but gravity is now opposed by buoyancy, so the effective weight is

m g (1 - \frac{ρ _{w}}{ρ})

tan \frac{θ}{2} = \frac{F / K}{m g ( 1 - \frac{ρ _{w}}{ρ} )} = \frac{q ^{2}}{4 π ε _{0} K r ^{2} m g ( 1 - \frac{ρ _{w}}{ρ} )}

Since

θ

(and hence

r

) is unchanged, equating both expressions for

tan (θ /2)

K (1 - \frac{ρ _{w}}{ρ}) = 1 ⟹ K = \frac{1}{1 - \frac{1}{1.5}} = \frac{1}{1/3} = 3

JEE Main · 2024 · 01 Feb 2024 · Shift 1 · Integer answer