NTA Abhyas JEE Main2020PhysicsElectrostaticsPractice
Four particles, each of mass m and charge q , are held at the vertices of a square of the side a as shown in the figure. They are released at t = 0 and move under mutual repulsive forces. Speed of any particle when its distance from the centre of square doubles, is -
Options
- A1 4 π ε 0 q 2 m a 1 + 1 2 2 1 / 2
- B1 4 π ε 0 q 2 m a 1 / 2
- C1 4 π ε 0 q 2 m a 2 1 / 2
- D1 4 π ε 0 2 q 2 m a 2 1 + 1 2 2 1 / 2
Correct answer
A. 1 4 π ε 0 q 2 m a 1 + 1 2 2 1 / 2
Step-by-step solution
Potential energy, when rude of square is x is given by U = 4 K q 2 x + 2 K q 2 2 x U i + K i = U f + K f K q 2 a 4 + 2 + 0 = K q 2 2 a 4 + 2 + 4 × 1 2 m v 2 K q 2 a 4 + 2 1 - 1 2 = 2 m v 2 K q 2 2 a 2 m 4 + 2 = v = K q 2 m a 1 + 1 2 2