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Two identical balls each having density P are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle theta with vertical. Now both the balls are immersed in liquid. As a result the angle does not change. the density of the liquid is sigma. find the dielectric constant of the liquid .

Two identical balls each having density P are suspended from a common point by two insulating strings of equal length. Both the balls have equal mass and charge. In equilibrium each string makes an angle theta with vertical. Now both the balls are immersed in liquid. As a result the angle does not change. the density of the liquid is sigma. find the dielectric constant of the liquid .

Grade:12

2 Answers

Pushkar Aditya
71 Points
10 years ago
Let Theta=Q and sigma = S TcosQ= mg TsinQ= coulomb force between the balls Make a free body diagram and find tangent of theta. Now take the second case in which balls are immersed in liquid. In this case we would take upthrust(=vSg) acting on the ball. Now TcosQ= mg + vSg TsinQ=(coulomb force between the balls)/K K is the dielectric constant of liquid. Again find tangent of theta. Since angle does not change so equate the two values of tangent of theta. Answer would come out K= P/(P-S)
ankit singh
askIITians Faculty 614 Points
3 years ago

Hence the dielectric constant for liquid is K = ρ / (ρ-σ )

Explanation:

Each ball is in equilibrium under following three forces:

i. tension, ii. electric force and iii. Weight

Lami's theorem can be applied

Where, K= dielectric constant of liquid and W' = W − upthrust Applying Lami's theorem in vacuum

W / sin(90∘ + θ) = Fe / sin(180∘ − θ)

Or W cos θ = Fe sinθ  -------(1)

Similarly in liquid

W'/ (cos θ ) = Fe'/ (sin θ)

Dividing equation (1) by (2) we get.

W / W' = Fe/Fe'

K = W / (W - "upthrust")

(as Fe/Fe' = K) = (Vρg) / (Vρg - Vσg)

V = volume of ball) or

K = ρ / (ρ-σ )

Hence the dielectric constant for liquid is  K = ρ / (ρ-σ )

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