Problem 54 - Master Resource Book in JEE Main Physics

Question

Two particles of equal mass \(m\) and charge \(q\) are placed at a distance of \(16 \mathrm{~cm}\). They do not experience any force. The value of \(\frac{q}{m}\) is (a) \(\sqrt{\frac{\pi \varepsilon_{0}}{G}}\) (b) \(\sqrt{\frac{G}{\pi \varepsilon_{0}}}\) (c) \(\sqrt{4 \pi \varepsilon_{0}} G\) (d) \(l\)

Step-by-Step Solution

Answer

The value of \( \frac{q}{m} \) is \( \sqrt{4 \pi \varepsilon_0 G} \) (Option c).

Step 1: Understanding the Problem

We have two particles with equal mass \( m \) and charge \( q \) placed at a distance of 16 cm from each other. They do not experience any net force, meaning the gravitational force is balanced by the electric force.

Step 2: Expression for Gravitational Force

The gravitational force between two particles is given by \[ F_g = \frac{G m^2}{r^2} \]where \(G\) is the gravitational constant, \(m\) is the mass, and \(r\) is the distance between the particles. Substituting \( r = 0.16 \) m, we have \[ F_g = \frac{G m^2}{(0.16)^2} \].

Step 3: Expression for Electric Force

The electric force between two charged particles is given by \[ F_e = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} \]where \( \varepsilon_0 \) is the permittivity of free space, \( q \) is the charge, and \( r \) is the distance between the particles. Using \( r = 0.16 \) m, we have \[ F_e = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(0.16)^2} \].

Step 4: Setting Forces Equal

Since the particles do not experience any net force, the gravitational force must equal the electric force. Therefore, \[ \frac{G m^2}{(0.16)^2} = \frac{1}{4 \pi \varepsilon_0} \frac{q^2}{(0.16)^2} \].

Step 5: Simplifying the Equation

Cancel \( (0.16)^2 \) from both sides, resulting in \[ G m^2 = \frac{1}{4 \pi \varepsilon_0} q^2 \].

Step 6: Solving for \( \frac{q}{m} \)

Rearrange the equation to find \[ \frac{q^2}{m^2} = 4 \pi \varepsilon_0 G \].Taking the square root on both sides, we get \[ \frac{q}{m} = \sqrt{4 \pi \varepsilon_0 G} \].

Key Concepts

Gravitational Force

Gravitational force is a fundamental force of attraction between two masses. In this context, it acts between two particles with mass, say particle A and particle B. Isaac Newton's law of universal gravitation describes this force mathematically as: \[ F_g = \frac{G m_1 m_2}{r^2} \] Here:

• \( F_g \) is the gravitational force.
• \( G \) is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
• \( m_1 \) and \( m_2 \) are the masses of the particles.
• \( r \) is the distance between them.

Gravitational force is always attractive, meaning it always tries to pull the two masses together. Regardless of their charges, the gravitational force between two particles is influenced solely by their masses and the distance separating them.

Electric Force

Electric force arises between charged particles. It can be either attractive or repulsive, depending on the nature of the charges involved. Coulomb's law gives us a way to calculate this force as follows: \[ F_e = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2} \] Where:

• \( F_e \) is the electric force.
• \( \varepsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).
• \( q_1 \) and \( q_2 \) are the charges of the particles.
• \( r \) is the distance between the charges.

In our scenario, if both particles have the same type of charge (either both positive or both negative), they repel each other. Conversely, if one is positive and the other is negative, they attract. This principle helps explain the force interaction that occurs independently of gravitational forces.

Force Equilibrium

Force equilibrium occurs when the total forces acting on a body are balanced, resulting in no net force. This balance is essential in systems where forces must neutralize each other to maintain stability or a constant state. In our exercise, the particles are said to "not experience any force," which means the electric force exerted by the charges precisely cancels the gravitational force due to their masses. This can be expressed as:

• \( F_g = F_e \)

For equilibrium:

• If \( F_g \) rises, \( F_e \) must rise to match it, and vice-versa.
• The conditions \( G m^2/r^2 = q^2/(4 \pi \varepsilon_0 r^2) \) reflect this balance by equating the magnitude of gravitational and electric forces.

Equilibrium emphasizes the interaction between these forces, ensuring that the net force sums to zero, allowing for a constant, unchanging system state.

Charge to Mass Ratio

The charge to mass ratio, represented as \( \frac{q}{m} \), is a critical feature of charged particles, providing insight into their behavior under electric or magnetic fields. It is pivotal in fields like electromagnetism and particle physics. When dealing with electric and gravitational interactions, the charge to mass ratio helps understand how much electric force a charge experiences relative to the gravitational force due to its mass. In our problem, it was found by equating the forces and simplifying:

• \( G m^2 = \frac{q^2}{4 \pi \varepsilon_0} \)
• Solving for \( \frac{q}{m} \), we found: \[ \frac{q}{m} = \sqrt{4 \pi \varepsilon_0 G} \]

This result tells us the specific balance at which gravitational pull and electric push perfectly counteract each other. Knowing the charge to mass ratio allows for precise calculations in diverse applications involving electric and magnetic fields.