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

The correct answer is option (C): \(\frac{\lambda_0}{8\pi\epsilon_0R}\hat{i}\)
The magnitude of the field = \(dE=\frac{1}{4\pi\epsilon_0}\frac{rd\theta\times Q/(\pi r/2)}{r^2}=\frac{Q}{2\pi^2\epsilon_0r^2}d\theta\)
\(E=\int_{0}^{\frac{\pi}{2}}2dEsin\theta =2\int_{0}^{\frac{\pi}{2}}\frac{Q}{2\pi\epsilon_0r^2}sin\theta d \theta =\frac{Q}{\pi^2\epsilon_0r^2}\)=

Answer 2

Take PO as the x-axis and PA as the y-axis. Consider two small elements, EF and E'F', at an angular distance θ above and below the line PO, each with a width of dθ.
Field Calculation: .
The magnitude of the electric field at point P due to either element is: .
dE = (1 / 4πε₀) * (rdθ * Q / (π/2)) / r² = Q / (2π²ε₀r²) dθ.
Components: .
The components of the field along the line PO cancel each other out. The resultant field is along the line PA (y-axis). .
Total Field: .
The field at point P due to both elements is: .
E = ∫₀^π/2 2E sinθ dθ.
Solving this integral gives: .
E = Q / (π²ε₀r²)