Two identical balls each having density σ are suspended from a common ...
Problem: Two identical balls each having density σ 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 if the balls are suspended in a liquid of density ρ and the distance between the balls remains same, then what will be the dielectric constant of the liquid?
Solution:
Concepts:
- Electrostatics
- Buoyancy
- Equilibrium of forces
Assumptions:
- The system is in equilibrium in both cases
- The charge on each ball remains the same
- The length of the strings remains the same
- The liquid is incompressible
- The balls are fully submerged in the liquid
Steps:
1. Calculate the tension in the strings in air and in liquid
2. Calculate the buoyant force on each ball in the liquid
3. Equate the net force on each ball in air and in liquid to find the dielectric constant of the liquid
Step 1: Tension in strings
- In air: The tension in each string is equal to the weight of the ball, i.e., T = mg
- In liquid: The tension in each string is equal to the weight of the ball minus the buoyant force, i.e., T = (m - ρV)g, where V is the volume of each ball
Step 2: Buoyant force
- The buoyant force on each ball in the liquid is equal to the weight of the liquid displaced by the ball, i.e., Fb = ρVg, where V is the volume of each ball
Step 3: Equating forces
- In air: The net force on each ball is equal to the tension in the string minus the weight, i.e., mgcosθ = T - mg
- In liquid: The net force on each ball is equal to the tension in the string minus the weight plus the buoyant force, i.e., (m - ρV)gcosθ = T - (m - ρV)g + ρVgcosθ
- Equating the two forces, we get (m - ρV)gcosθ = T - (m - ρV)g + ρVgcosθ
- Simplifying, we get ρVgcosθ = (m - ρV)g
- Solving for ρ, we get ρ = m/(2Vcosθ - 2mσ/ρ)
Answer: The dielectric constant of the liquid is ρ = m/(2Vcosθ - 2mσ/ρ).
Two identical balls each having density σ are suspended from a common ...
Tan@in air= F/mg
tan@in liquid =F*/mg-v(rho)g m=v (sigma)g
@in air =@in liq F* =F/E ^
E^=sigma-rho/sigma