Problem 137 - Master Resource Book in JEE Main Physics

Question

A thin semi-circular ring of radius \(r\) has a positive charge \(q\) distributed uniformly over it. The net field \(\mathbf{E}\) at the centre \(O\), is (a) \(\frac{q}{4 \pi^{2} \varepsilon_{0} r^{2}} \hat{\mathbf{j}}\) (b) \(-\frac{q}{4 \pi^{2} \varepsilon_{0} r^{2}} \hat{\mathbf{j}}\) (c) \(-\frac{q}{2 \pi^{2} \varepsilon_{0} r^{2}} \hat{\mathbf{j}}\) (d) \(\frac{q}{2 \pi^{2} \varepsilon_{0} r^{2}} \hat{\mathbf{j}}\)

Step-by-Step Solution

Answer

The answer is (c) \(-\frac{q}{2\pi^2 \varepsilon_0 r^2} \hat{\mathbf{j}}\).

Step 1: Understand the system

We have a semi-circular ring with radius \( r \) and total positive charge \( q \) distributed uniformly over its arc. We aim to find the net electric field \( \mathbf{E} \) at the center \( O \) of the semi-circle.

Step 2: Consider the charge element

Take a small charge element \( dq \) at an angle \( \theta \) from the diameter of the semi-circle. Since the charge is uniformly distributed, the linear charge density \( \lambda \) is \( \frac{q}{\pi r} \). The charge element is \( dq = \lambda r d\theta = \frac{q}{\pi} d\theta \).

Step 3: Determine the electric field due to a charge element

The electric field \( d\mathbf{E} \) at the center due to \( dq \) is radial and has magnitude \( dE = \frac{k_e dq}{r^2} \) where \( k_e = \frac{1}{4\pi \varepsilon_0} \) is Coulomb's constant. Thus, \( dE = \frac{k_e q}{\pi r^2} d\theta \).

Step 4: Resolve into components

Each \( dE \) can be resolved into horizontal and vertical components, \( dE_x = dE \sin \theta \) and \( dE_y = dE \cos \theta \). Summing over the semi-circle, all \( dE_x \) components cancel due to symmetry, leaving only \( dE_y \) contributing to the net field.

Step 5: Calculate net vertical component

The vertical component is \( dE_y = \frac{k_e q}{\pi r^2} \cos \theta \, d\theta \). Integrate from \( \theta = 0 \) to \( \theta = \pi \):\[ E_y = \int_{0}^{\pi} \frac{k_e q}{\pi r^2} \cos \theta \, d\theta = \frac{k_e q}{\pi r^2} \left[ \sin \theta \right]_{0}^{\pi} = \frac{2k_e q}{\pi r^2}. \]

Step 6: Express final result

The direction of \( E_y \) is negative \( y \)-direction because the \( y \)-components point upwards for the top half and downward for the bottom (stronger) half. Thus, the net electric field at the center is:\[ \mathbf{E} = -\frac{q}{2\pi^2 \varepsilon_0 r^2} \hat{\mathbf{j}}. \]

Step 7: Choose the correct option

Comparing this result with the options given, the correct answer is (c) \(-\frac{q}{2\pi^2 \varepsilon_0 r^2} \hat{\mathbf{j}}\).

Key Concepts

Charge Distribution

In physics, the concept of charge distribution is crucial for understanding how electric fields are created and behave in different systems. A charge distribution refers to how electric charge is spatially spread over a particular region or object. It can be distributed along a line, over a surface, or throughout a volume. In the exercise we are discussing, the charge is distributed along a semi-circular ring, which means it is an example of a linear charge distribution.
When a charge is

• Spread uniformly, as in our example, it means that each portion of the arc carries the same amount of charge per unit length.
• Described mathematically, this uniformity is expressed as linear charge density, denoted by the symbol \( \lambda \).

Understanding how a charge is distributed is essential because it directly determines the characteristics of the electric field generated by the charge. Knowing the distribution helps us apply mathematical techniques to compute fields like the electric field at specific points, such as the center of the semi-circle in this problem.

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics, describing how two point charges interact with each other. This law states that the force between two charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. The formula for Coulomb's Law is:\[F = k_e \frac{|q_1 q_2|}{r^2}\]where

• \( F \) is the magnitude of the force
• \( k_e = \frac{1}{4\pi\varepsilon_0} \) is Coulomb's constant
• \( q_1 \) and \( q_2 \) are the electric charges
• \( r \) is the distance between the charges

This exercise involves calculating the electric field rather than a force, which can be derived from Coulomb's Law. By considering an infinitesimally small charge element \( dq \) along the semi-circular ring, we can apply Coulomb's Law to find the electric field \( d\mathbf{E} \) due to \( dq \).
The field can then be determined at any point, like the center of our semi-circle, using this approach.

Integration of Electric Field

To find the total electric field due to a distributed charge, we use the method of integration. In this exercise, the goal is to determine how each infinitesimal charge element contributes to the electric field at a specific point, and then sum all contributions. This process involves calculating the electric field contribution from small sections of charge and integrating those contributions over the entire charge distribution.
Here's how it works:

• Every small charge element \( dq \) creates an electric field at the point of interest.
• Each field has a radial component with magnitude given by \( dE = \frac{k_e dq}{r^2} \).
• Due to symmetry, the horizontal components cancel out when summed, simplifying the calculation to only vertical components.
• This leads to an integral of the vertical components \( dE_y = \frac{k_e q}{\pi r^2} \cos \theta \, d\theta \), resulting in the total electric field.

Integration helps manage complex charge distributions and allows us to predict the resulting electric field with precision, as we did to find the net electric field at the center of our semi-circular ring.

Linear Charge Density

Linear charge density is a way to describe how charge is distributed along a line, like our semi-circular ring. It is defined as the amount of charge per unit length along the line and is denoted by the symbol \( \lambda \). In terms of formula, when a total charge \( q \) is distributed uniformly over a length \( L \), linear charge density is given by:\[\lambda = \frac{q}{L}\]For our semi-circular ring, the arc length is \( \pi r \), leading to:\[\lambda = \frac{q}{\pi r}\]

• This density allows us to express a small charge element \( dq \) as \( dq = \lambda r d\theta = \frac{q}{\pi} d\theta \).
• Using linear charge density simplifies the application of Coulomb's Law over a continuous distribution of charge.
• It's important in calculating quantities like electric fields, as it provides a straightforward link between distributed charge and its physical effects.

In summary, understanding linear charge density is essential for dealing with distributed charges, enabling calculations such as those needed to find the electric field at the center of this semi-circle.