Problem 15 - Master Resource Book in JEE Main Physics

Question

A charged oil drop falls with terminal velocity \(v_{0}\) in the absence of electric field. An electric field \(E\) keeps it stationary. The drop acquires charge \(3 q\), it starts moving upwards with velocity \(v_{0}\). The initial charge on the drop is (a) \(\frac{q}{2}\) (b) \(q\) (c) \(\frac{3 q}{2}\) (d) \(2 q\)

Step-by-Step Solution

Answer

The initial charge on the drop is \(\frac{3q}{2}\), option (c).

Step 1: Understanding the Problem

A charged oil drop initially falls with a terminal velocity \(v_0\) due to gravity. When an electric field \(E\) is applied, it balances the gravitational force, keeping the drop stationary. When the charge is increased to \(3q\), the oil drop starts moving upwards with the same velocity \(v_0\). We need to determine the initial charge of the drop that allowed it to stay stationary in the electric field.

Step 1: Force Balance in Stationary State

When the drop is stationary due to the electric field \(E\), the electric force \(qE\) on the charge \(q\) equals the gravitational force \(mg\) plus the viscous drag force (when moving with \(v_0\), but zero when at rest). So, we have the equality: \[ qE = mg \]

Step 2: Analyze Movement with New Charge

When the charge becomes \(3q\), the net force is now directed upwards, and the oil drop moves with velocity \(v_0\) due to the net upward force being equal to the original downward gravitational force and upward electric force. So, \[ 3qE - mg = mg \] This simplifies to \[ 3qE = 2mg \]

Step 3: Solve for Initial Charge

From the stationary state equation, we have \(qE = mg\). Replacing \(mg\) in the equation for the moving state, we have: \[ 3qE = 2(qE) \] Dividing both sides by \(E\), we get: \[ 3q = 2q \] Thus, the initial charge \(q\) is \(\frac{3q}{2}\).

Step 5: Conclusion

Given the analysis and calculations, the initial charge on the drop is \(\frac{3q}{2}\). Therefore, the correct option is (c) \(\frac{3q}{2}\).

Key Concepts

Electric Force

An electric force arises when a charged object interacts with an electric field. This force can either repel or attract other charged objects depending on the type of charges involved. In the case of the charged oil drop, the electric force helps counteract the gravitational pull on the drop.
To calculate the electric force acting on a charge, we use the formula:

• Electric Force = Charge (q) × Electric Field (E)

The direction and magnitude of the electric force is directly related to the charge. If the charge is positive, the force will be in the same direction as the electric field. Conversely, if the charge is negative, the force will be opposite to the electric field direction. In the exercise, the initial electric force that balances gravity on the oil drop makes it stay stationary, meaning that the electric force exactly matches the gravitational force.

Gravitational Force

Gravitational force is the attractive force exerted by the Earth on objects due to their mass. It is responsible for pulling objects towards the Earth's center. The formula for gravitational force (weight) is:

• Gravitational Force = Mass (m) × Acceleration due to gravity (g)

In terms of the charged oil drop, this force governs its motion when there is no electric field present. The oil drop initially moves downward with terminal velocity due to this gravitational pull. When the electric field is introduced, it balances out this gravitational force, allowing us to calculate the conditions needed for equilibrium and analyze the effects of adding more charge.

Electric Field

An electric field is a region around a charged object where other charged objects experience a force. The field is represented by the symbol 'E' and is measured in volts per meter (V/m). It gives both the direction and magnitude of the force a charge would experience if it were placed in that field.
The electric field is essential in the given exercise, as it acts against the gravitational force to keep the oil drop stationary. The initial situation is one of balance with the electric field exactly equating the gravitational force, allowing us to find the value of the initial charge. When the field maintains the drop stationary, it means the initial charge matches the forces precisely.

Charge Calculation

Charge calculation is critical in understanding how different forces act on objects based on their charge. For the charged oil drop, knowing its charge allows us to predict how it interacts with an electric field and how its motion changes with different charge values.
The exercise shows a scenario where the initial charge, under a stationary situation, is calculated using balance equations:

• In stationary mode: \( qE = mg \)
• With increased charge \( (3q) \): \( 3qE = 2mg \)

By solving these equations, we find the initial charge on the drop. These calculations help in scenarios where understanding the balance between electric and gravitational forces is crucial. In this case, the charge is ultimately found to be \( \frac{3q}{2} \), matching the problem's requirement for stationary balance.