Problem 131 - Master Resource Book in JEE Main Physics

Question

Two positive charges of magnitude \(q\) are placed at the end of a side 1 of a square of side \(2 a\). Two negative charges of the same magnitude are kept at the other corners. Starting from rest, if a charge \(Q\) moves from the middle of side 1 to the centre of square, its kinetic energy at the centre of square is (a) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q Q}{a}\left(1-\frac{1}{\sqrt{5}}\right)\) (b) zero (c) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q Q}{a}\left(1+\frac{1}{\sqrt{5}}\right)\) (d) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{2 q Q}{a}\left(1-\frac{2}{\sqrt{5}}\right)\)

Step-by-Step Solution

Answer

The kinetic energy at the center is given by (a) \(\frac{1}{4\pi \varepsilon_0} \frac{2qQ}{a}\left(1-\frac{1}{\sqrt{5}}\right)\).

Step 1: Identify Given Setup

We have a square with side length \(2a\) and four charges. At the corners of one side, we have two positive charges \(+q\) each, and at the opposite corners, we have two negative charges \(-q\) each. A charge \(Q\) moves from the middle of one side to the center of the square.

Step 2: Calculate Initial Location Potential

The potential energy at the starting point, which is the middle of one side, is calculated due to the four corner charges. Let this point be midway between two positive charges. Hence, the potential at this position due to each positive charge is \(V_+ = \frac{1}{4\pi \varepsilon_0} \frac{q}{a}\) since it is located directly at distance \(a\) from each. The potentials due to the two negative charges are calculated similarly, and both are at distance \(\sqrt{(2a)^2 + a^2} = \sqrt{5}a\) giving \(V_- = \frac{1}{4\pi \varepsilon_0} \frac{-q}{\sqrt{5}a}\). Thus, the initial potential energy \( U_i = Q (2V_+ + 2V_-) = \frac{1}{4\pi \varepsilon_0} \cdot Q \left( \frac{2q}{a} - \frac{2q}{\sqrt{5}a} \right) \).

Step 3: Calculate Center Location Potential

When the charge \(Q\) is at the center of the square, all four corners are equidistant from \(Q\) with distance \(\sqrt{2}a\). Therefore, the potential due to any charge is \(V_c = \frac{1}{4\pi \varepsilon_0} \cdot \frac{q}{\sqrt{2}a}\), giving a total potential \( U_f = Q (2 V_c - 2 V_c) = 0 \), because there are two positive and two negative contributions that cancel out.

Step 4: Calculate Kinetic Energy at Center

Initially, \(Q\) is at rest; hence, starting kinetic energy is zero and potential energy was \(U_i\). When \(Q\) reaches the center, the potential energy is \(U_f = 0\). By conservation of energy, initial potential energy has converted entirely into kinetic energy. Therefore, the kinetic energy at the center is equal to the initial potential energy: \(K = U_i = \frac{1}{4\pi \varepsilon_0} \cdot Q \left( \frac{2q}{a} - \frac{2q}{\sqrt{5}a} \right) = \frac{1}{4\pi \varepsilon_0} \cdot \frac{2qQ}{a} \left( 1 - \frac{1}{\sqrt{5}} \right) \).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics that describes the force between two point charges. This law states that the electrostatic force between any two charges is directly proportional to the product of their magnitudes and inversely proportional to the square of the distance between them. Mathematically, it is given by:

• Equation: \( F = \frac{1}{4 \pi \varepsilon_0} \frac{|q_1 q_2|}{r^2} \)
• \(\varepsilon_0\): Permittivity of free space.
• \(q_1\) and \(q_2\): The magnitudes of the two charges.
• \(r\): The distance between the charges.

This law is crucial for understanding how charges interact and form the basis for calculating forces in various configurations. In our exercise, Coulomb's Law helps in determining the potential energy contributed by each charge in the system, which subsequently influences calculations related to energy conversions.

Electric Potential Energy

Electric potential energy is the energy a charge possesses due to its position within an electric field. This energy is derived from the work done to bring a charge to a specific point from infinity. For a point charge, the electric potential \(V\) due to another point charge is defined as:

• Equation: \( V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r} \)
• \(q\): The charge causing the potential \(V\).
• \(r\): The distance of the point from the charge \(q\).

In our scenario, the potential energy calculations involve evaluating contributions from multiple charges, each affecting the overall potential seen by the moving charge \(Q\). By summing the individual potentials due to each charge at a given point, we calculate the total potential energy at that position, which is essential for analyzing the initial and final energy states of the system.

Conservation of Energy

The conservation of energy principle asserts that energy in a closed system is constant, merely changing forms from one type to another without any loss. In electrostatics, we're often concerned with conversions between potential energy and kinetic energy. When the charge \(Q\) moves in an electric field, its potential energy transforms into kinetic energy as work is done on it.In the provided exercise, the charge starts from rest, meaning its initial kinetic energy is zero. As the charge moves towards the center of the square, potential energy decreases, and kinetic energy increases while the system fulfills:

• Initial Energy \(=\) Final Energy
• Potential Energy Loss \(=\) Kinetic Energy Gain

Hence, by accounting for the initial potential energy calculated using Coulomb’s law, we establish the final kinetic energy at the center of the square. This beautifully underscores the transformative nature of energy within electrostatic fields.