A fixed non-conducting smooth track is in the shape of a quarter circle of radius R in a vertical plane. A small metal ball A is fixed at the bottom of the track. Another identical ball B, which is free to move, is placed in contact with ball A. A charge Q is given to ball A which gets equally shared by the two balls. Ball B gets repelled and ultimately comes to rest in its equilibrium position, where its radius vector makes an angle θ = 60° with vertical. Mass of ball B is m. Find charge Q that was given to the balls.
(a) R√πε₀mg
(b) 2R√πε₀mg
(c) 3R√πε₀mg
(d) 4R√πε₀mg
Answer (1)
To solve this problem, we need to analyze the forces acting on ball B when it reaches its equilibrium position after being repelled by ball A due to electrostatic force.
Understand the Setup:
Ball A is fixed at the bottom of the quarter circle track, and initially, it has a charge Q.
When ball A is given the charge Q, the charge is equally shared between balls A and B, so each ball gets a charge of 2 Q .
Ball B is repelled and comes to rest at an angle θ = 6 0 ∘ with the vertical.
Forces on Ball B at Equilibrium:
At equilibrium, the electrostatic force between the two balls and the gravitational force acting on ball B are in balance.
The electrostatic force is given by Coulomb's Law as: F e = r 2 k ( 2 Q ) 2 where k = 4 π ε 0 1 and r = R sin θ because it is on the circular track.
The gravitational force component pulling ball B backwards along the track is: F g = m g sin θ
Establish Equilibrium Condition:
At equilibrium, the component of the electrostatic force along the track must balance the gravitational component: R 2 sin 2 θ k ( 2 Q ) 2 = m g sin θ
Solve for Q:
By substituting k = 4 π ε 0 1 and sin θ = 2 3 (since θ = 6 0 ∘ ) into the equilibrium condition, we have: 4 π ε 0 1 ⋅ ( R ⋅ 2 3 ) 2 ( 2 Q ) 2 = m g ⋅ 2 3
Simplifying gives: ( 2 Q ) 2 = 4 π ε 0 m g R
Solve for Q : Q = 2 R π ε 0 m g
Thus, the charge Q that was initially given to the balls is 2 R π ε 0 m g , corresponding to option (b).
