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JEE Advanced 2016 · Paper 2, Physics question 18

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PARAGRAPH 2

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)

A closed cylindrical chamber with conducting top and bottom plates and many small balls resting on the bottom plate. The top plate connects through an ammeter A to the negative terminal of a high-voltage source HV; the bottom plate connects to its positive terminal.

The average current in the steady state registered by the ammeter in the circuit will be

Choose one option
Answer and worked solution

Official answer: D

Charge per ball. q∝V0q\propto V_0 (equipotential with the plate).

Field. E∝V0E\propto V_0, so the acceleration qE/m∝V02qE/m\propto V_0^2.

Trip time. t=2h/a∝1V0t=\sqrt{2h/a}\propto\dfrac1{V_0}.

Average current. Each trip carries charge ∝ qq, so
I∝qt∝V0⋅V0=V02.I\propto\frac qt\propto V_0\cdot V_0=V_0^2.

(D)

Topics:Drift velocity and currentElectric field
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