Problem 50 - Master Resource Book in JEE Main Physics

Question

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

Step-by-Step Solution

Answer

Option (b) \( \frac{\lambda}{2 \pi \varepsilon_0 a} \) is correct.

Step 1: Understand the Geometry and Problem

We have a semicircular arc with a radius 'a'. It is uniformly charged with a linear charge density \( \lambda \). This means the charge per unit length along the arc is \( \lambda \). We need to find the electric field at the center of the semicircle, O.

Step 2: Define the Expression for Electric Field

The electric field generated by a small charge element \( dq \) at the center can be expressed as \( dE = \frac{k \, dq}{r^2} \) where \( k = \frac{1}{4\pi\varepsilon_0} \) and \( r = a \). Here, \( dq = \lambda \, ds \), with \( ds \) being a small element of arc length.

Step 3: Use Symmetry and Integrate

Consider the semicircle's symmetry: each segment at \( \theta \) has a counterpart at \( -\theta \). As a result, the horizontal components of their fields cancel, while vertical components (along the y-axis) add up. The vertical component of \( dE \) is \( dE_y = dE \sin \theta \). Thus, \( dE_y = \frac{k \lambda \sin \theta \ ds}{a^2} \). Using \( ds = a \, d\theta \), we integrate from \( -\frac{\pi}{2} \) to \( \frac{\pi}{2} \): \( E_y = \int_{-\pi/2}^{\pi/2} \frac{\lambda}{4\pi\varepsilon_0 \ a} \sin \theta \, d\theta \).

Step 4: Calculate the Integral

The integral \( \int_{-\pi/2}^{\pi/2} \sin \theta \, d\theta = 2 \). Substituting, \( E_y = \frac{\lambda}{4\pi\varepsilon_0 \ a} \times 2 = \frac{\lambda}{2\pi\varepsilon_0 \ a} \).

Step 5: Choose the Correct Option

After calculation, we observe that the correct expression for the net electric field at the center is \( \frac{\lambda}{2 \pi \varepsilon_0 a} \), which matches option (b).

Key Concepts

Symmetry in Physics

Physics often relies on symmetry to simplify complex problems. In the case of an electric field from a charged semicircular arc, symmetry is key. A semicircle has mirror symmetry about the vertical axis passing through the center.
This means for each small element of charge on one side of the semicircle, there is another element directly symmetrical on the opposite side.

• The horizontal components of their electric fields cancel each other out, because they have equal magnitude but opposite direction.
• However, the vertical components are along the same direction and therefore add up.

By understanding this symmetry, we simplify the task of calculating the resultant electric field to dealing with only the vertical components, thus making it much easier to integrate over the entire arc.
This approach highlights the power of symmetry in simplifying calculations not just in this problem, but across many areas in physics.

Linear Charge Density

Linear charge density, represented as \( \lambda \), is a measure of the amount of electric charge per unit length along an object.
In the context of a charged semicircular arc, knowing the linear charge density helps us determine the total charge distributed along the arc easily.
Since it is uniformly charged, \( \lambda \) remains constant, simplifying other calculations because every infinitesimal section of the arc can be treated identically in terms of charge.

• We express each small element of charge, \( dq \), as \( dq = \lambda \, ds \), where \( ds \) is a small segment of the arc length.

For a semicircle with radius \( a \), the length of the arc (which is half the circumference of a full circle) is \( \pi a \).
Therefore, the total charge on the semicircle can be calculated as the product of the linear charge density and the arc length, or \( \lambda \pi a \).
This fundamental understanding helps in setting up the integral for the electric field, grounding the problem in terms of charge distribution.

Integration in Physics

Integration is an essential tool in physics for summing continuous distributions of quantities, like charge and electric fields.
In our semicircular arc problem, it allows us to calculate the total electric field produced at the center by summing up infinitesimal contributions from each small section of the arc.

• First, consider a small arc element with an associated electric field contribution \( dE \).
• We then express \( dE \) as a function of \( \theta \), where \( \theta \) is the angle in polar coordinates, simplifying the problem to a manageable integral over \( \theta \).

Following the integration from \( -\pi/2 \) to \( \pi/2 \), we take advantage of calculus to account for continuous change.
Such integration tells us about directionality (as shown by integrating \( \sin \theta \)), effectively capturing how the varying orientation of \( dq \) pieces adds to the net field at the center.
Integrating \( \int \sin \theta \, d\theta \) from \( -\pi/2 \) to \( \pi/2 \) yields \( 2 \), demonstrating that full understanding through integration is crucial in evaluating fields from continuous charge distributions.