Problem 126 - A Complete Resource Book in Physics for JEE Main

Question

A charge \(q_{1}\) is placed at the centre of a spherical conducting shell of radius \(R\). Conducting shell has a total charge \(q_{2} .\) Electrostatic potential energy of the system (A) \(\frac{q_{1}^{2}+2 q_{1} q_{2}}{8 \pi \varepsilon_{0} R}\) (B) \(\frac{q_{2}^{2}+2 q_{1} q_{2}}{8 \pi \varepsilon_{0} R}\) (C) \(\frac{q_{1}^{2}+q_{1} q_{2}}{4 \pi \varepsilon_{0} R}\) (D) \(\frac{q_{2}^{2}+q_{1} q_{2}}{4 \pi \varepsilon_{0} R}\)

Step-by-Step Solution

Answer

The electrostatic potential energy of the system is given by option (C): \(U = \frac{q_1^2 + q_1q_2}{4\pi\varepsilon_0R}\).

Step 1: Calculate the electrostatic potential energy between the charges

Since we are calculating the potential energy between the central charge \(q_1\) and the charges on the shell, we can use the formula \(U = \frac{Kq_1q_2}{R}\), where \(K = \frac{1}{4\pi\varepsilon_0}\) and the distance between the charges is \(R\). Now, substituting the value of \(K\), we get: \(U = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1q_2}{R}\) From this equation, we see that none of the given options match it exactly. Thus, we need to analyze further to find the correct answer.

Step 2: Analyze the energy distribution

Since the entire energy of the system is due to an interaction between charge \(q_1\) and charge \(q_2\), we can infer that the energy of the system is divided between the energies due to each individual charge. The energy from charge \(q_1\) can be represented as \(\frac{1}{2}\cdot \frac{q_1^2}{4\pi\varepsilon_0R}\). Similarly, the energy from charge \(q_2\) can be represented as \(\frac{1}{2}\cdot \frac{q_2^2}{4\pi\varepsilon_0R}\). Therefore, the total energy can be written as: \(U = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1^2}{4} + \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_2^2}{4} + \frac{1}{4\pi\varepsilon_0}\cdot\frac{2q_1q_2}{R}\)

Step 3: Choose the correct answer

Combining the terms in the equation obtained in Step 2, we get: \(U = \frac{1}{4\pi\varepsilon_0}\cdot\frac{q_1^2 + q_1q_2 + q_2^2}{R}\) Comparing this equation with the given options, we see that option (C) matches the expression: \(U = \frac{q_1^2 + q_1q_2}{4\pi\varepsilon_0R}\) Hence, the electrostatic potential energy of the system is given by option (C).

Key Concepts

Coulomb's Law

To fully grasp the concept of electrostatic potential energy in this exercise, let's delve into Coulomb's Law, a fundamental principle in electrostatics which describes the force between two point charges. According to this law, the force (F) between two charges is proportional to the product of the magnitudes of the charges \(q_1\ and \q_2\), and inversely proportional to the square of the distance \(r\) between them. Mathematically, this is expressed as:

• \( F = K \cdot \frac{|q_1 q_2|}{r^2} \)

Here, \(K\) represents Coulomb's constant, calculated as \(\frac{1}{4\pi \varepsilon_0}\), where \(\varepsilon_0\) is the permittivity of free space. In our exercise, the charges \(q_1\) and \(q_2\) interact at the center of a conducting shell, so their interaction is simplified at a distance \(R\). The equation for potential energy, as seen in our solution, utilizes Coulomb's law to approximate interactions with:

• \( U = \frac{K \cdot q_1 q_2}{R} \)

Understanding this principle is crucial, as it establishes the basis for calculating forces and potential energy in electrostatics.

Conducting Shell

In electrostatics, a conducting shell is crucial when analyzing electrostatic potential energy. It is a hollow conductor which can completely surround other charges. When a charge is placed inside, as in our exercise with \(q_1\) at the center of the spherical shell, the electric field inside the conductor is zero. This happens because the charges on the shell rearrange themselves to cancel out any internal field, ensuring equilibrium.The charges on the conducting shell will distribute uniformly due to repulsive forces if the shell itself is similarly charged or neutral. This redistribution impacts the potential energy calculations because now we only consider interactions across the shell interface. In our problem, the conducting shell of radius \(R\) holds a charge \(q_2\). When calculating potential energy, one only considers the source and test charges' separation distance \(R\), beyond which all internal field effects are nullified.This property is particularly useful because once the shell achieves equilibrium, the potential inside the shell due to itself is constant (zero), simplifying many complex calculations.

Electrostatics Formulas

Electrostatics deals heavily with formulas that describe the relationships between electric fields, forces, and energies. To determine the correct electrostatic potential energy in our exercise, we use a blend of these formulas.

• Potential Energy Formula: This defines the energy between charges \(q_1\ and \q_2\) as \(U = \frac{K \cdot q_1 q_2}{R}\), showing dependence on both charges and their separation.
• Energy Distribution: The energy components from the individual charges in relation to their configurations. Here, we calculated contributions as \(\frac{1}{2}\frac{q_1^2}{4\pi\varepsilon_0R}\) and \(\frac{1}{2}\frac{q_2^2}{4\pi\varepsilon_0R}\) for respective charge energies.
• Superposition Principle: This principle states that the total energy is a sum of all individual energy interactions. As illustrated, \(U = \frac{1}{4\pi\varepsilon_0}\frac{q_1^2+q_1q_2+q_2^2}{R}\) combines all energy effects within the system.

These formulas and principles together facilitate a comprehensive understanding of electrostatic potential energy, aiding in solving problems like the one given while also deepening general knowledge in electrostatics.