Sign In

JEE Advance - Physics (2016 - Paper 2 Offline - No. 17)

Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h. Now, a high voltage source (HV) connected across the conducting plates such that the bottom plate is at +V0 and the top plate at $$-$$V0. Due to their conducting surface, the balls will get charge, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)

JEE Advanced 2016 Paper 2 Offline Physics - Electrostatics Question 29 English
Which one of the following statement is correct?
The balls will execute simple harmonic motion between the two plates
The balls will bounce back to the bottom plate carrying the same charge they went up with
The balls will stick to the top plate and remain there
The balls will bounce back to the bottom plate carrying the opposite charge they went up with

Explanation

The distance between the two plates is h. The potential of the bottom plate is V0 and that of the top plates is $$-$$V0. The electric field between the plates is E = 2V0/h (directed upwards). The radius of each ball is r (<< h). Let m be the mass and C be the capacitance of each ball.

JEE Advanced 2016 Paper 2 Offline Physics - Electrostatics Question 29 English Explanation

When the ball touches the bottom plate, it gets a positive charge q = CV0 (we assume that the charge transfer is instantaneous). This positively charged ball experiences an upward force, F = qE = 2CV$$_0^2$$/h, which accelerates the ball upwards. Since the force is constant, the ball cannot do SHM (for SHM, the force should be proportional to the displacement and directed towards the centre).

When the ball hits the top plate, it transfers the positive charge to the plate and gets negative charge q = $$-$$CV0. This negatively charged ball again experience a force F = qE (downward) and starts accelerating downwards. Thus, the ball keeps moving between the bottom and the top plates carrying a charge +q upwards and $$-$$q downwards.

Comments (0)

Advertisement