JEE Challenger
More from Electrostatic Potential and Capacitance
PhyMedAdv 2016 • Paper 2

Behavior of Conducting Balls in High Voltage Electric Field

Comprehension Passage

Consider an evacuated cylindrical chamber of height hh 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≪hr \ll h. Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0+V_0 and the top plate at −V0-V_0. Due to their conducting surface, the balls will get charged, 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)

Passage Diagram 1

Identify the correct statement regarding the motion and charging state of the spherical balls:

Options

A

The balls will stick to the top plate and remain there

B

The balls will bounce back to the bottom plate carrying the same charge they went up with

C

The balls will bounce back to the bottom plate carrying the opposite charge they went up with

Correct
D

The balls will execute simple harmonic motion between the two plates

Topics & Concepts

Electrostatic Potential and CapacitanceElectric PotentialElectric FieldCapacitors

Step-by-Step Solution

To determine the correct statement regarding the motion and charging state of the conducting spherical balls, let us analyze the physical process step-by-step:

  1. Initial State and Charging at the Bottom Plate:

    • The bottom plate is maintained at a potential of +V0+V_0, while the top plate is at −V0-V_0.
    • Initially, the conducting balls rest on the bottom plate. Due to electrical contact, free electrons flow from the balls to the bottom plate, causing the balls to acquire a positive charge q>0q > 0 until they become equipotential with the bottom plate.

  2. Direction of Electric Field and Force:

    • The electric field E⃗\vec{E} inside the parallel plate setup points from the higher potential region (bottom plate) to the lower potential region (top plate). Therefore, E⃗\vec{E} is directed vertically upwards.
    • The electrostatic force on a positively charged ball is given by: F⃗=qE⃗\vec{F} = q\vec{E} Since q>0q > 0 and E⃗\vec{E} points upwards, the net force is directed upwards, repelling the balls from the bottom plate towards the top plate.

  3. Interaction at the Top Plate:

    • As the balls reach the top plate, they collide with it. Although the coefficient of restitution is zero (e=0e = 0) due to the soft material (meaning their velocity temporarily drops to zero), they make electrical contact with the top plate at potential −V0-V_0.
    • Upon contact, charge transfer occurs instantaneously: the balls lose their positive charge and acquire a negative charge −q′-q' to match the potential of the top plate.

  4. Return Motion:

    • With a negative charge −q′-q', the electrostatic force acting on the ball becomes: F⃗′=(−q′)E⃗\vec{F}' = (-q')\vec{E}
    • Since E⃗\vec{E} points upwards, the force F⃗′\vec{F}' acts vertically downwards (away from the top plate and towards the bottom plate).
    • This downward electrostatic force immediately accelerates the balls back down toward the bottom plate, carrying a charge opposite to the charge they carried when moving upward.

Thus, the balls will bounce back to the bottom plate carrying the opposite charge to what they went up with.

Correct Option: (C)

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