Problem 16 - 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) \(\frac{\lambda}{2 \pi \varepsilon_{0}^{a^{2}}}\) (b) \(\frac{\lambda}{4 \varepsilon_{0} a}\) (c) \(\frac{\lambda^{2}}{4 \pi \varepsilon_{0} a}\) (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 have a semi-circular arc with radius \(a\) and a uniform charge distribution. The charge per unit length is \(\lambda\). We need to find the electric field at the center of this semicircle.

Step 2: Identify the Electric Field Contribution from a Small Segment

Consider a small element of the arc, an infinitesimally small angle \(d\theta\). The length of this element is \(ds = a \, d\theta\) and its charge is \(dq = \lambda \, ds = \lambda a \, d\theta\).

Step 3: Determine the Electric Field from the Small Segment

The electric field \(dE\) due to charge \(dq\) at the center is directed radially outward from the arc. The magnitude of \(dE\) is given by Coulomb's law: \(dE = \frac{1}{4 \pi \varepsilon_0} \frac{dq}{a^2}\). Substitute \(dq\) to get \(dE = \frac{1}{4 \pi \varepsilon_0} \frac{\lambda a \, d\theta}{a^2}\).

Step 4: Resolve Electric Field into Components

The electric field due to each segment can be resolved into a vertical component \(dE_y\) and a horizontal component \(dE_x\). By symmetry, the horizontal components cancel out. Thus, we focus on \(dE_y = dE \sin\theta\).

Step 5: Integrate Over the Semicircle

The net vertical electric field \(E_y\) is obtained by integrating \(dE_y\) over the entire semicircle from \(-\pi/2\) to \(\pi/2\): \[ E_y = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4 \pi \varepsilon_0 a} \sin\theta \, d\theta \]Evaluating this integral, we find\[ E_y = \frac{\lambda}{4 \pi \varepsilon_0 a} \left[ -\cos\theta \right]_{-\pi/2}^{\pi/2} = \frac{\lambda}{4 \pi \varepsilon_0 a} [0 - (\- 0)] = \frac{2\lambda}{4 \pi \varepsilon_0 a} \].

Step 6: Simplify the Expression

After simplification, the net electric field at the center is \(E = \frac{\lambda}{2 \pi \varepsilon_0 a}\) which corresponds to option (d).

Key Concepts

Coulomb's Law

Coulomb's Law forms the foundation for understanding electric fields. This principle tells us how two charged objects interact, regardless of whether they have like or unlike charges. The law states that the force \( F \) between two charges \( q_1 \) and \( q_2 \) is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance \( r \) between them. This can be mathematically expressed as: \[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where \( k = \frac{1}{4 \pi \varepsilon_0} \) is Coulomb's constant, and \( \varepsilon_0 \) is the permittivity of free space.
In the context of electric fields, Coulomb's Law informs us that a charge \( dq \), however small, will create an electric field element \( dE \) that radiates outward. In exercises involving charge distributions like a semi-circular arc, the law helps break down how each unit of charge contributes to the field at a specific point, such as the center.

Charge Distribution

When dealing with continuous charge distributions, it's crucial to understand how the charges are spread out along an object. In this exercise, we have a semi-circular arc, uniformly charged with a charge per unit length \( \lambda \). Uniform distribution means that \( \lambda \) is constant all along the arc, giving us a straightforward basis for calculating fields.
The total charge of any small segment on the arc, with infinitesimally small angle \( d\theta \), can be described as \( dq = \lambda \cdot a \cdot d\theta \). Here, \( a \) is the radius of the arc, giving each segment length \( ds = a \cdot d\theta \).
To calculate the electric field caused by these charges at the center, you treat each segment's charge as creating its own field. Because of symmetry across the semicircle, complexity reduces significantly. The horizontal components of the electric field created by each segment cancel each other out, leaving only vertical components.

Electric Field Components

Breaking down the electric field into components helps clarify which parts contribute to the total field at a particular point. For a uniformly charged arc, the electric field at the center due to a small charged segment \( dq \) has components that need to be understood separately. The field from \( dq \) points radially outward.
The radial field \( dE \) can be split into horizontal \( dE_x \) and vertical \( dE_y \) components. Symmetrically, the \( dE_x \) components from one half of the arc cancel out those from the other half due to their equal magnitude and opposite directions.
The only contributing parts left are the vertical components \( dE_y \), which add up coherently. We integrate these components over the entire semicircle from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \). The integration of this sine-based function allows us to find the net electric field strength at the center of the semicircle, simplifying to \[E = \frac{\lambda}{2 \pi \varepsilon_0 a}\].
This confirms the contributions from the individual field components add up to match option (d) in the original exercise. All calculations show how understanding individual electric field components enables accurate predictions about the total field in complex charge distributions.