Problem 73 - Master Resource Book in JEE Main Physics

Question

Two insulated charged conducting spheres of radii \(20 \mathrm{~cm}\) and \(15 \mathrm{~cm}\), respectively and having an equal charge of \(10 \mu \mathrm{C}\) are connected by a copper wire and then they are separated. Then (a) both spheres will have equal charges (b) surface charge density on the \(20 \mathrm{~cm}\) sphere will be greater than that on the \(15 \mathrm{~cm}\) sphere (c) surface charge density on the \(15 \mathrm{~cm}\) sphere will be greater than that on the \(20 \mathrm{~cm}\) sphere (d) surface charge density on the two spheres will be equal

Step-by-Step Solution

Answer

(c) Surface charge density on the 15 cm sphere is greater than that on the 20 cm sphere.

Step 1: Understand the Problem Statement

Two conducting spheres with radii 20 cm and 15 cm are initially charged with 10 μC each. They are connected by a copper wire (allowing charge to flow and reach equilibrium) and then separated. We need to determine the relationship between their surface charge densities afterwards.

Step 2: Total Charge Conservation

The total charge is conserved during the process. Initially, both spheres together have a charge of \( Q_1 + Q_2 = 10 \, \mu \mathrm{C} + 10 \, \mu \mathrm{C} = 20 \, \mu \mathrm{C} \).

Step 3: Equilibrium Condition

When connected by a wire, charges will redistribute until the electric potential on both spheres is the same. The potential \( V \) of a sphere with charge \( Q \) and radius \( R \) is given by \( V = \frac{kQ}{R} \), where \( k \) is Coulomb's constant.

Step 4: Setting Potentials Equal

Since the potentials are the same for both spheres, \( \frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \). Substituting the radii, we get: \( \frac{Q_1}{20} = \frac{Q_2}{15} \).

Step 5: Solving the Equation

Solve for one of the charges using the relation \( Q_1 = \frac{20}{15}Q_2 \). Combine with the total charge equation: \( Q_1 + Q_2 = 20 \, \mu \mathrm{C} \). Substitute \( Q_1 = \frac{4}{3}Q_2 \) in this equation to find \( Q_2 \).

Step 6: Finding Individual Charges

Substitute \( Q_1 \) in terms of \( Q_2 \) into the total charge conservation equation: \[ \frac{4}{3}Q_2 + Q_2 = 20 \, \mu \mathrm{C} \]. Solve, \( \frac{7}{3}Q_2 = 20 \), so \( Q_2 = \frac{60}{7} \, \mu \mathrm{C} \approx 8.57 \, \mu \mathrm{C} \). Thus, \( Q_1 \approx 11.43 \, \mu \mathrm{C} \).

Step 7: Calculate Surface Charge Densities

Surface charge density \( \sigma \) is given by \( \sigma = \frac{Q}{4\pi R^2} \). Calculate \( \sigma_1 \) for the 20 cm sphere and \( \sigma_2 \) for the 15 cm sphere. Compare the densities: \( \sigma_1 \approx 2.27 \, \mu \mathrm{C/m}^2 \) and \( \sigma_2 \approx 3.05 \, \mu \mathrm{C/m}^2 \).

Step 8: Conclusion

Since \( \sigma_2 > \sigma_1 \), the surface charge density on the 15 cm sphere is greater than that on the 20 cm sphere.

Key Concepts

Charge Distribution

When dealing with conductors, an important concept is how charge distributes itself on a surface when equilibrium is reached. Let's imagine two spheres, each charged to 10 μC, connected by a wire. Once connected, electrons flow until each sphere reaches the same electric potential. It's key to remember that while the charges might change on each sphere, the total charge remains the same at 20 μC. The distribution is such that the potential, defined by the equation \[ V = \frac{kQ}{R} \], equalizes. Because potential depends on both charge \( Q \) and radius \( R \), each sphere ends up with a different amount of charge, preserving total charge. Therefore, charge distribution depends directly on the geometry and size of each object involved.

Electric Potential

Electric potential is a critical aspect when dealing with electrostatics. It's the measure of the potential energy per unit charge at a point in a field. For a sphere, it's determined by the formula \( V = \frac{kQ}{R} \). Here, \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere. When two charged spheres are connected, their charges redistribute. Redistribution continues until both spheres have the same electric potential. This condition ensures energy equilibrium. Although charges \( Q_1 \) and \( Q_2 \) change, the potentials \( V_1 \) and \( V_2 \) come into harmony, equating the sums \( \frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \). This fundamental step in electrostatics reflects the nature of electric fields to stabilize and create uniform potential across connected conductors. It emphasizes that potential, not the charge directly, is the conserved element across the system.

Surface Charge Density

Surface charge density \( \sigma \) is the amount of charge per unit area on a surface. For a sphere, it's given by \( \sigma = \frac{Q}{4\pi R^2} \). In this context, it represents how tightly packed the charge is on the sphere's surface. After connecting two spheres and allowing charges to redistribute, each sphere achieves a unique charge \( Q \). Given their different radii \( R_1 = 20 \text{ cm} \) and \( R_2 = 15 \text{ cm} \), they end up with different surface charge densities. Despite equalizing potentials, the smaller sphere, with lesser area, ends up having a higher surface charge density. This is illustrated by \( \sigma_1 = 2.27 \, \mu \mathrm{C/m^2} \) for the 20 cm sphere, and \( \sigma_2 = 3.05 \, \mu \mathrm{C/m^2} \) for the 15 cm sphere, demonstrating that surface charge density is not only dependent on charge but significantly influenced by the surface area.