Question

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

A. Proportional to $V_0^2$
B. Proportional to the potential ${V_0}$
C. Zero
D. Proportional to $V_0^{1/2}$

Answer 1

In this case, we will take the coefficient of restitution to be zero as the balls are light in weight. The electric field generated inside the chamber can be taken as a parallel plate capacitor. We will find the relation of total charge and potential difference. After that we will find the relation of time and potential difference, using the equation of distance from one of the equations of the laws of motion.

Formula used:
The acceleration of a particle in an electric field is given by:
$a = \dfrac{{Q \times E}}{m}$
Where,
$a$ indicates the acceleration.
$Q$ indicates the charge of the particle.
$E$ indicates the electric field.
$m$ indicates the mass of the particle.

The electric field is given by:
$E = \dfrac{V}{d}$
Where,
$V$ indicates potential difference.
$d$ indicates distance.

Complete step by step answer:
In the given question, we are supplied with the following data:
There is a container which contains a number of balls. These balls are charged by a high potential. The balls are considered to be lightweight and they are coated with some conducting material. After the electric current is allowed to pass, the ball will also be positively charged along with the plate and they will be repelled by the plate.

Let us proceed to solve the problem. We assume the height of the container to be $h$ .The total charge on the balls is $Q$ .The voltage applied on the balls is ${V_0}$ .We know that the charge on the balls is directly proportional to the voltage applied on the balls. Mathematically, we can write:
Q \propto {V_0} \\\ \Rightarrow Q = k{V_0} \\\
When the balls will bounce upwards, they will cover a distance of $h$ .
So, we can now write:
h = \dfrac{1}{2}a{t^2} \\\ \Rightarrow h = \dfrac{1}{2} \times \dfrac{{Q \times E}}{m} \times {t^2} \\\ \Rightarrow h = \dfrac{1}{2} \times \dfrac{{Q \times {V_0}}}{{m \times h}} \times {t^2} \\\ \Rightarrow h = \dfrac{1}{2} \times \dfrac{{k{V_0} \times {V_0}}}{{m \times h}} \times {t^2} \\\ \Rightarrow h = \dfrac{1}{2} \times \dfrac{{kV_0^2}}{{m \times h}} \times {t^2} \\\ \Rightarrow {t^2} = \dfrac{{2{h^2}m}}{{kV_0^2}} \\\ \Rightarrow t = \dfrac{{h\sqrt {2m} }}{{\sqrt k {V_0}}} \\\
Now, all the quantities are constant except $t$ and ${V_0}$ .
So, the relation becomes:
$\Rightarrow t \propto \dfrac{1}{{{V_0}}}$
Again, we can write for the current:
${I_{{\text{avg}}}} \propto \dfrac{Q}{t} \\\ \Rightarrow {I_{{\text{avg}}}} \propto \dfrac{{k{V_0}}}{{k \times \dfrac{1}{{{V_0}}}}} \\\ \therefore {I_{{\text{avg}}}} \propto V_0^2$
Hence, current is proportional to $V_0^2$ .

The correct option is A.

Note: Remember that an electric field is an elegant way of characterizing a charging system's electrical environment. At any point in the space around a charging device, the electric field reflects the force that a positive unit test charge will feel if mounted at that point.