Problem 76 - Master Resource Book in JEE Main Physics

Question

Two insulated metal spheres of radii \(10 \mathrm{~cm}\) and \(15 \mathrm{~cm}\) charged to a potential of \(150 \mathrm{~V}\) and \(100 \mathrm{~V}\) respectively, are connected by means of a metallic wire. What is the charge on the first sphere? (a) 2 esu (b) 4 esu (c) 6 esu (d) 8 esu

Step-by-Step Solution

Answer

The charge on the first sphere after connection is 1200 esu.

Step 1: Understand the Basics

When two conductors are connected by a metallic wire, the potential (voltage) across both will equalize due to charge redistribution. Initially, the spheres have potentials of 150 V and 100 V.

Step 2: Final Potential Consideration

Let the final common potential be \( V_f \). Since the spheres will be at the same potential upon connection, both spheres will share their total charge equally across their respective capacities.

Step 3: Capacitance of a Sphere

The capacitance \( C \) of a sphere is given by the formula \( C = R \) in electrostatic units (esu), where \( R \) is the radius. Therefore, for the first sphere \( C_1 = 10 \text{ cm} \) and for the second sphere \( C_2 = 15 \text{ cm} \).

Step 4: Initial Charges on the Spheres

The initial charge \( Q_1 = C_1 \times 150 = 10 \times 150 = 1500 \text{ esu} \) and \( Q_2 = C_2 \times 100 = 15 \times 100 = 1500 \text{ esu} \).

Step 5: Final Charge Distribution

Using the principle of conservation of charge, the total charge \( Q_{total} = Q_1 + Q_2 = 1500 + 1500 = 3000 \text{ esu} \). After equalization, the charge will be distributed: \( V_f = \frac{Q_{total}}{C_1 + C_2} = \frac{3000}{10 + 15} = 120 \text{ V} \).

Step 6: Charge on First Sphere After Connection

The charge \( Q_1 \) residing on the first sphere will be \( Q'_1 = V_f \times C_1 = 120 \times 10 = 1200 \text{ esu} \).

Key Concepts

Charge Distribution

When two charged conductors, like our metal spheres, are connected by a wire, their charges redistribute until both are at the same potential.
This is a key concept in electrostatics known as charge distribution.
Here's how it works:

• Each conductor initially has its own charge and potential.
• Upon connection, charges will flow between the conductors.
• The flow continues until both conductors reach the same potential. This is due to the principle that charge will move to equalize potential between connected conductive objects.

For the spheres in the exercise, they start at different potentials of 150 V and 100 V.
Connection through the wire allows charges to flow and equalize, resulting in a common potential, which was calculated as 120 V.
Understanding charge distribution helps us predict how initial charges will rearrange when conductors are connected, facilitating further solutions in electrostatic problems.

Conductors and Potential

Conductors are materials that allow free flow of electric charge.
This property causes charges to move until the potential is uniform over the surface of the conductor.
Important things about conductors:

• Inside a perfect conductor, the electric field is zero when at electrostatic equilibrium.
• The potential within the conductor is constant; any differing potential would cause a redistribution of charges until uniformity is achieved.

In the provided exercise, the spheres are initially not at the same potential.
When they are connected, they act as a single conductor and adjust their potentials to one uniform value.
After the connection, the final potential can be calculated using the principle of conservation of charge, which governs how the potential becomes uniform.

Capacitance of a Sphere

Capacitance is a measure of a conductor's ability to store charge, largely dependent on its geometry and surroundings.
For isolated spheres, capacitance in electrostatic units (esu) is equivalent to the radius of the sphere (R).
Helpful points about sphere capacitance:

• For a sphere with radius R, capacitance C in esu is equal to R.
• This simplicity means for sphere 1 we get C = 10 esu, and for sphere 2, C = 15 esu based on their radii.

The exercise example calculates the initial and final charges using these capacitance values:
The initial charges are derived from multiplying capacitance by their respective potentials.
After charge redistribution, the charges on each sphere can be recalculated using the equalized final potential and respective capacitance.
Understanding this concept is essential for accurately calculating how charges are affected under changing electrical conditions.