Problem 102 - Master Resource Book in JEE Main Physics

Question

27 identical drops of mercury are charged simultaneously to the same potential of \(10 \mathrm{~V}\) each. Assuming drops to be spherical, if all the charged drops are made to combine to form one larger drop, then the potential of larger drop would be (a) \(45 \mathrm{~V}\) (b) \(135 \mathrm{~V}\) (c) \(270 \mathrm{~V}\) (d) \(90 \mathrm{~V}\)

Step-by-Step Solution

Answer

The potential of the larger drop is 90 V, option (d).

Step 1: Understand the Problem

We have 27 identical drops with the same potential of 10 V each. We need to find the potential of a single, larger drop formed by combining all these smaller drops.

Step 2: Use the Formula for the Potential of a Sphere

The potential \( V \) of a spherical drop is given by \( V = \frac{kQ}{R} \), where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the drop.

Step 3: Calculate the Total Charge

For each small drop, \( V = \frac{kQ}{r_0} = 10 \). Therefore, \( Q = 10r_0/k. \) The total charge \( Q_t \) of the large drop is the sum of the charges of all small drops. Thus, \( Q_t = 27 \times Q = 27 \times \frac{10r_0}{k} = \frac{270r_0}{k}. \)

Step 4: Calculate the Radius of the Large Drop

The volume of the large drop is equal to the sum of volumes of all small drops. Thus, \( \frac{4}{3}\pi R^3 = 27 \times \frac{4}{3}\pi r_0^3 \). Solving for \( R \), we get \( R = 3r_0. \)

Step 5: Calculate the Potential of the Large Drop

Using the formula \( V = \frac{kQ}{R} \), substitute \( Q_t \) and \( R \) to find the potential of the large drop: \[ V = \frac{k \cdot \frac{270r_0}{k}}{3r_0} = \frac{270}{3} = 90\, \text{V}. \]

Step 6: Choose the Correct Answer

Based on the calculation, the potential of the larger drop is 90 V. Therefore, the correct answer is \( (d)\, 90 \,\text{V}. \)

Key Concepts

Potential of a Sphere

The concept of the potential of a sphere is essential in understanding electrics in spherical objects. The potential, denoted as \( V \), refers to the electric potential energy per unit charge at a point in space due to an electric field. For a spherical object, such as the drops in our exercise, this potential is determined using the formula:

• \( V = \frac{kQ}{R} \)

where:

• \( k \) is Coulomb's constant, \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \)
• \( Q \) is the total electric charge of the sphere
• \( R \) is the radius of the sphere

In this scenario, when we combine 27 smaller drops into one large drop, the challenge is to determine this new potential. To do so, we rely on the changes in total charge and radius while using the same formula to ascertain the new potential.

Charge Distribution

The way charge is distributed on a sphere affects its overall electric behavior. For a sphere, the charge is uniformly distributed over its surface.In our problem, each small drop has the same potential of 10 V. Thus, each drop's charge can be derived using the relation \( Q = \frac{10r_0}{k} \). To find out the total charge for the large drop formed by combining 27 smaller drops, we consider:

• The charge on each small drop, \( Q = \frac{10r_0}{k} \)
• The total charge, \( Q_t = 27 \times Q = \frac{270r_0}{k} \)

This approach shows how charges accumulate when individual charges coalesce, affecting the result of the potential for the larger drop.

Combination of Charged Drops

When multiple charged drops combine, their volumes contribute to the formation of a single large drop. The charge remains conserved, meaning the total charge of the large drop is the sum of all individual charges.Using the equation for the volume of a sphere, we understand:

• The volume of one small drop is \( \frac{4}{3}\pi r_0^3 \)
• For 27 small drops, their total volume is \( 27 \times \frac{4}{3}\pi r_0^3 \)
• The volume of the large drop is \( \frac{4}{3}\pi R^3 \), where \( R = 3r_0 \) when solved

Thus, by substituting into the formula, we ensure both mass and charge conservation, key elements to finding new potential.

Coulomb's Law

Coulomb's Law is a foundational principle used to calculate electrostatic forces between charged objects. Expressed as \( F = \frac{k |q_1 q_2|}{r^2} \), it plays a vital role in determining how charges interact.Although the exercise primarily involves potentials rather than forces directly, Coulomb's constant (\( k \)) is pivotal here. It helps establish the electric potential of the sphere by factoring into \( V = \frac{kQ}{R} \).Understanding Coulomb's Law is essential to grasp the influence of charges and how changes in configuration, like combining smaller drops into a large one, alter potential distributions and outcomes.