Problem 55 - Master Resource Book in JEE Main Physics

Question

A semi-circular arc of radius \(a\) is charged uniformly and the charge per unit length is \(\lambda\). The electric field at its centre is (a) \(\frac{\lambda}{2 \pi \varepsilon_{0}+^{2}}\) (b) \(\frac{\lambda}{4 \varepsilon_{0} a}\) (c) \(\frac{\lambda^{2}}{2 \pi e_{0} a}\) (d) \(\frac{\lambda}{2 \pi \varepsilon_{0} a}\)

Step-by-Step Solution

Answer

The electric field at the center is (d) \(\frac{\lambda}{2\pi \varepsilon_0 a}\).

Step 1: Understand the Problem

We are given a semi-circular arc of radius \(a\) with a uniform linear charge density \(\lambda\). We need to find the electric field at the center of curvature of the semi-circle.

Step 2: Symmetry and Coordinates

Since the semi-circular arc is symmetric about the y-axis, the horizontal components of electric field from symmetrically opposite elements cancel out. You only need to consider the vertical component of the electric field.

Step 3: Electric Field Due to an Element

Consider a small element of arc \(d\theta\) at an angle \(\theta\). The charge on this small element \(dq = \lambda a d\theta\). The electric field due to this small charge element at the center is \(dE = \frac{1}{4\pi \varepsilon_0} \cdot \frac{dq}{a^2}\).

Step 4: Components of Electric Field

Decompose the element's electric field \(dE\) into x- and y-components: only the y-components \(dE_y = dE \cos(\theta)\) will contribute to the net electric field, because \(E_x\) cancels due to symmetry.

Step 5: Integrate to Find Total Electric Field

Integrate \(dE_y\) from \(\theta = -\pi/2\) to \(\theta = \pi/2\). \[ E = \int_{-\pi/2}^{\pi/2} \frac{\lambda a d\theta}{4\pi \varepsilon_0 a^2} \cos \theta = \frac{\lambda}{4\pi \varepsilon_0 a} \int_{-\pi/2}^{\pi/2} \cos \theta \, d\theta \]The integral of \(\cos \theta\) over \([-\pi/2, \pi/2]\) is 2. Hence,\[ E = \frac{\lambda}{2\pi \varepsilon_0 a} \]

Step 6: Match with Given Options

Check the calculated electric field expression with the given options. The calculated electric field, \( \frac{\lambda}{2\pi \varepsilon_0 a} \), matches with option (d). Thus, this is the correct answer.

Key Concepts

Semi-Circular Arc

In electromagnetics, a semi-circular arc is a curved line forming half of a circle. Imagine slicing a circle into two equal halves, and you get a semi-circular arc. This arc has a radius, often referred to as \(a\). When dealing with the electric field due to such an arc, it's crucial to analyze its geometrical properties.
One key detail is its symmetry, which simplifies calculations. If you picture the arc vertically aligned, the right and left sides mirror each other along the vertical axis. This symmetry plays a significant role, especially in vector quantities like electric fields. Components in opposing directions may cancel each other, simplifying the overall impact.

• Geometry: Semi-circle with radius \(a\).
• Symmetry: Mirror image along the vertical axis.

This mirroring is why in many exercises involving electric fields on arcs or circles, only certain components need consideration. Understanding the geometry of a semi-circular arc is foundational for further exploring the electric field at its center.

Uniform Charge Distribution

When we say a semi-circular arc has a uniform charge distribution, this means that the charge is spread evenly along the length of the arc. The charge per unit length, denoted as \(\lambda\), remains constant throughout the arc.
Think of it as evenly spreading butter on a slice of bread, where every point on the slice gets an equal amount. This uniformity allows us to easily calculate the charge on any segment of the arc.
The formula for charge on a small segment is straightforward:

• \(dq = \lambda a d\theta\)

Here, \(dq\) is the infinitesimal charge on a tiny segment of the arc; \(d\theta\) represents a small angle subtended by the segment at the center. This constant \(\lambda\) makes integration possible since you can apply it across the entire arc without adjusting for varying charge densities. Hence, uniform charge distribution simplifies calculations in electrostatics, allowing predictive and consistent results.

Electrostatics

Electrostatics is a branch of physics focusing on electric charges at rest. Unlike moving charges that produce magnetic fields, static ones influence only electric fields.
In our exercise, the semi-circular arc's charges are static. This means they create an electric field around them while staying in place. Calculating this field involves considering many aspects including geometry, charge distribution, and symmetry.
Electrostatics uses several fundamental laws to understand these effects, such as Gauss's Law and sometimes more directly, Coulomb's Law. Because dealing with continuous charge distributions can be complex, we often break it into infinitesimally small charges and sum up their effects using calculus.

• Key Concept: Charges at rest, creating steady electric fields.
• Important Laws: Gauss and Coulomb's laws.

Electrostatics provides a rich foundation for understanding interactions between charged objects in a system, such as the forces and fields they generate.

Coulomb's Law

Coulomb's Law describes the electrostatic interaction between charged objects. It essentially tells us the force between two point charges depends on their magnitudes and the distance between them.
The formula for Coulomb’s Law is:
\[ F = \frac{k \, |q_1 \, q_2|}{r^2} \]
where \(F\) is the force, \(k\) the electrostatic constant, \(q_1\) and \(q_2\) the charges, and \(r\) the distance between them.
In our exercise, we're dealing with a continuous charge distribution. Thus, we use a modified approach where small charge elements \(dq\) contribute to the electric field \(dE\). The principle remains similar: calculate the field due to each small charge element, and integrate across the arc.

• Direct Use: Calculating fields from infinitesimal charges.
• Integration: To compute total field contributions.

Understanding Coulomb’s Law is pivotal for grasping how charges interact, influencing both the forces they exert and the fields they produce.