Animated Solution for Physics - Electrostatics: Comprehension Passage
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 te 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)
Question 1:
Which one of the following statement is correct?
Select Answer:
Question 2:
The average current in the steady state registered by the ammeter in the circuit will be
Select Answer:
Visualized Solution
Understanding the Setup
Bottom plate potential: +V0
Top plate potential: −V0
Charge on a Single Ball
Potential of the ball at the bottom plate: V=V0
Capacitance of a spherical ball: C=4πε0r
Charge acquired: q=CV=4πε0rV0
Electric Field in the Chamber
Potential difference between plates: ΔV=V0−(−V0)=2V0
Distance between plates: h
Uniform electric field: E=hΔV=h2V0
Force on the Ball
Electrostatic force: F=qE
Acceleration of the Ball
Acceleration: a=mqE=m(4πε0rV0)(h2V0)
a=mh8πε0rV02
Time of Flight
Kinematic equation: h=21at2
Time taken to reach the top plate: t=a2h
Substituting a: t=8πε0rV022hmh=V0h4πε0rm
Average Current
Current is the rate of charge transfer: I=tnq
Substitute q∝V0 and t∝V01
I∝1/V0V0∝V02
Analyzing the Motion (Question 4)
At the top plate (−V0), the ball acquires a negative charge: q′=−4πε0rV0.
The electric field E is upwards. Force on negative charge is downwards.
The ball is repelled back to the bottom plate.
Since F is constant, a is constant. Motion is oscillatory, not SHM (F∝−x).
The Way Forward
Consider the effect of air resistance on the terminal velocity and average current.
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The Sigma Insight: Capacitance and Capacitors
Solution Diagram
Analyzing the Setup
Imagine this cylindrical chamber. The bottom plate is at a high positive potential, +V0, and the top plate is at a negative potential, −V0. The little conducting balls resting at the bottom will acquire a positive charge, and because like charges repel, they will be shot upwards towards the top plate!
Let's focus on just one ball. Since it's touching the bottom plate, its potential becomes V0. We know the capacitance of a small sphere is 4πε0r. So, the charge q it picks up is simply its capacitance times the potential, which gives us:
q=4πε0rV0
Now, what's the electric field driving this ball? The chamber acts like a giant parallel plate capacitor. The total potential difference is V0−(−V0), which is 2V0. Dividing this by the height h gives us the uniform electric field E inside the chamber:
E=h2V0
The Master Equation
With the charge and electric field known, we can find the electrostatic force acting on the ball. The force F is simply the charge q times the electric field E. This upward force is what propels the ball towards the top plate.
By Newton's second law, the acceleration a is this force divided by the mass m. Substituting our expressions for q and E, we get:
a=mqE=m(4πε0rV0)(h2V0)=mh8πε0rV02
Notice carefully that the acceleration is directly proportional to the square of V0.
The ball starts from rest and travels a distance h. Using the second equation of motion, h=21at2, we can solve for the time of flight, t:
t=a2h=8πε0rV022hmh=V0h4πε0rm
When we plug in our acceleration, notice that V02 is in the denominator inside the square root. So, the time t is inversely proportional to V0.
Final Calculation
Finally, let's find the average current. Current is the total charge transported per unit time. For n balls, it's n times q divided by t:
I=tnq=V0h4πε0rmn(4πε0rV0)
Since the charge q is proportional to V0, and the time t is inversely proportional to V0, dividing them gives us a current that is proportional to V02! A beautiful result.
I∝V02
Analyzing the Motion
Let's also answer the conceptual question. When a ball hits the top plate, the coefficient of restitution is zero, meaning it stops momentarily. However, it now touches the negative plate and acquires a negative charge!
The upward electric field now exerts a downward force on this negative charge, pushing it back to the bottom plate. So, it bounces back carrying the opposite charge. And since the force is constant, not proportional to displacement, it's an oscillatory motion, but not Simple Harmonic Motion.