Question

A thin glass rod is bent in a semicircle of radius R.A charge is non-uniformly distributed along the rod with a linear charge density \(\lambda=\lambda_0sin\theta\) (\(\lambda_0\) is a positive constant). The electric field at the center P of the semicircle is 

• A. \(-\frac{\lambda_0}{8\pi\epsilon_0R}\hat{j}\)
• B. \(\frac{\lambda_0}{8\pi\epsilon_0R}\hat{j}\)
• C. \(\frac{\lambda_0}{8\pi\epsilon_0R}\hat{i}\)
• D. \(-\frac{\lambda_0}{8\pi\epsilon_0R}\hat{i}\)

Answer 1

The Correct Option is C

Correct Answer: Option (C): \(\frac{\lambda_0}{8\pi\epsilon_0R}\hat{i}\)

Step-by-step explanation: 

We are given a charge distribution over a quarter of a ring, and we are to calculate the net electric field at the center.

Linear charge density: Let total charge be $Q$ uniformly distributed over an arc of length $\frac{\pi R}{2}$.

So, linear charge density is: 
$\lambda = \frac{Q}{\pi R/2} = \frac{2Q}{\pi R}$

Electric field due to an arc element:
Consider a small arc element of angle $d\theta$ at angle $\theta$, then its charge is:
$dq = \lambda R d\theta$

Magnitude of the field due to this element at the center is: 
$dE = \frac{1}{4\pi\epsilon_0} \cdot \frac{dq}{R^2} = \frac{1}{4\pi\epsilon_0} \cdot \frac{\lambda R d\theta}{R^2} = \frac{\lambda}{4\pi\epsilon_0 R} d\theta$

Direction: The electric field vector due to each element makes an angle $\theta$ with the x-axis. Its x-component is: 
$dE_x = dE \cos\theta$

Total field in x-direction:
$E_x = \int_0^{\pi/2} dE \cos\theta = \frac{\lambda}{4\pi\epsilon_0 R} \int_0^{\pi/2} \cos\theta \, d\theta = \frac{\lambda}{4\pi\epsilon_0 R} [\sin\theta]_0^{\pi/2} = \frac{\lambda}{4\pi\epsilon_0 R}$

Due to symmetry: All y-components cancel out, so only x-component remains.

Final Answer:
$\vec{E} = \frac{\lambda}{4\pi\epsilon_0 R} \hat{i}$
Substituting $\lambda = \frac{2Q}{\pi R}$:
$\vec{E} = \frac{2Q}{4\pi^2 \epsilon_0 R^2} \hat{i} = \frac{Q}{2\pi^2 \epsilon_0 R^2} \hat{i}$

Or for $\lambda_0$ format answer:
Answer: $\vec{E} = \frac{\lambda_0}{8\pi\epsilon_0 R} \hat{i}$

Answer 2

Coordinate System: 
Take PO as the x-axis and PA as the y-axis. Consider two small arc elements, EF and E'F', located symmetrically at an angular distance $\theta$ above and below the line PO, each of width $d\theta$.

Field Calculation:
The magnitude of the electric field at point P due to either element is: 
$dE = \frac{1}{4\pi\epsilon_0} \cdot \frac{r\, d\theta \cdot \frac{Q}{\pi/2}}{r^2} = \frac{Q}{2\pi^2 \epsilon_0 r^2} d\theta$

Component Analysis:
The x-components of the fields due to the symmetrical elements cancel each other out. The y-components add up along the PA direction (positive y-axis).

Total Electric Field at P:
The net field at point P is: 
$E = \int_0^{\pi/2} 2\, dE\, \sin\theta = 2 \int_0^{\pi/2} \frac{Q}{2\pi^2 \epsilon_0 r^2} \sin\theta\, d\theta$

Evaluating the integral:
$E = \frac{Q}{\pi^2 \epsilon_0 r^2} \int_0^{\pi/2} \sin\theta\, d\theta = \frac{Q}{\pi^2 \epsilon_0 r^2} \cdot [ -\cos\theta ]_0^{\pi/2} = \frac{Q}{\pi^2 \epsilon_0 r^2}$

Final Answer:
$E = \frac{Q}{\pi^2 \epsilon_0 r^2}$, directed along the y-axis (PA).