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

Step 1: Understanding the Concept:
A uniformly charged semi-circular ring produces an electric field at its center directed along the axis of symmetry. The magnitude depends on the total charge and the radius.
Step 2: Key Formula or Approach:
The electric field at the center of a half-ring is: \[ E = \frac{2k\lambda}{R} \] where \(\lambda = \frac{q}{\pi R}\) is the linear charge density, \(R\) is the radius, and \(k = 9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\).
Step 3: Detailed Explanation:
Substitute \(\lambda\) into the electric field equation: \[ E = \frac{2k}{R} \left( \frac{q}{\pi R} \right) = \frac{2kq}{\pi R^2} \] We are given \(E = 100 \, \text{N/C}\) and \(R = 10 \, \text{cm} = 0.1 \, \text{m}\). Rearrange the formula to solve for \(q\): \[ q = \frac{E \pi R^2}{2k} \] Substitute the values: \[ q = \frac{100 \times \pi \times (0.1)^2}{2 \times 9 \times 10^9} \] \[ q = \frac{100 \times \pi \times 0.01}{18 \times 10^9} = \frac{\pi}{18 \times 10^9} \] \[ q = \frac{\pi}{18} \times 10^{-9} \, \text{C} \]
Step 4: Final Answer:
The charge on the ring is \(\frac{\pi}{18} \times 10^{-9} \, \text{C}\).