Question

Electric field due to a half ring at the center is \(100\,\text{N/C}\). Find the charge on the ring. Radius of the ring is \(10\,\text{cm}\).

• A. \( \dfrac{\pi}{9} \times 10^{-9}\,\text{C} \)
• B. \( \dfrac{\pi}{27} \times 10^{-9}\,\text{C} \)
• C. \( \dfrac{\pi}{18} \times 10^{-9}\,\text{C} \)
• D. \( \dfrac{\pi}{36} \times 10^{-9}\,\text{C} \)

Answer 1

The Correct Option is C

Concept: For a uniformly charged semicircular ring, the electric field at the center is \[ E=\frac{2k\lambda}{r} \] where

• \(k=\dfrac{1}{4\pi\varepsilon_0}=9\times10^9\,\text{N m}^2/\text{C}^2\)
• \(\lambda\) = linear charge density
• \(r\) = radius of the semicircle

Also, \[ \lambda=\frac{q}{\pi r} \] since the length of a semicircle is \( \pi r \). Step 1: Substitute \( \lambda = \dfrac{q}{\pi r} \) into the electric field formula.} \[ E=\frac{2k}{r}\left(\frac{q}{\pi r}\right) \] \[ E=\frac{2kq}{\pi r^2} \]
Step 2: Substitute the given values.} \[ E=100\,\text{N/C}, \quad k=9\times10^9, \quad r=10\,\text{cm}=10\times10^{-2}\,\text{m} \] \[ 100=\frac{2\times 9\times10^9 \times q}{\pi (10\times10^{-2})^2} \] \[ 100=\frac{18\times10^9 q}{\pi \times 10^{-2}} \]
Step 3: Solve for the charge \(q\).} \[ q=\frac{100 \times \pi \times 10^{-2}}{18\times10^9} \] \[ q=\frac{\pi}{18}\times10^{-9}\,\text{C} \] Final Result \[ q=\frac{\pi}{18}\times10^{-9}\,\text{C} \]