Question

Figure shows a rod AB, which is bent in a 120\(^{\circ}\) circular arc of radius R. A charge (\(-Q\)) is uniformly distributed over rod AB. What is the electric field \(\vec{E}\) at the centre of curvature O? 

 

• A. \( \frac{3\sqrt{3} Q}{8 \pi^2 \varepsilon_0 R^2} (\hat{i}) \)
• B. \( \frac{3\sqrt{3} Q}{8 \pi^2 \varepsilon_0 R^2} (-\hat{i}) \)
• C. \( \frac{3\sqrt{3} Q}{8 \pi \varepsilon_0 R^2} (\hat{i}) \)
• D. \( \frac{3\sqrt{3} Q}{16 \pi^2 \varepsilon_0 R^2} (\hat{i}) \)

Hint:

Memorizing the formula \( E = \frac{2k\lambda}{R} \sin(\alpha) \) for the electric field of a charged arc is a significant time-saver in exams. Remember that \(2\alpha\) is the total angle subtended by the arc. Always use symmetry to determine the direction of the net field before starting the calculation.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We need to calculate the electric field vector at the center of a circular arc that has a uniform charge distribution. The total charge is \(-Q\) and the arc subtends an angle of 120\(^{\circ}\).
Step 2: Key Formula or Approach:
The electric field at the center of a uniformly charged circular arc of radius \(R\), subtending a total angle \(2\alpha\) at the center, is given by:
\[ E = \frac{2k\lambda}{R} \sin(\alpha) \] where \( k = \frac{1}{4\pi\varepsilon_0} \) and \( \lambda \) is the linear charge density. The direction of the field is along the angle bisector.
Step 3: Detailed Explanation:
1. Setup and Symmetry:
The arc is symmetric about the x-axis, extending from \( \theta = -60^{\circ} \) to \( \theta = +60^{\circ} \). The total angle subtended is \( 2\alpha = 120^{\circ} \), so \( \alpha = 60^{\circ} \).
Due to this symmetry, the y-components of the electric field from infinitesimal charge elements will cancel out. The net electric field will be along the x-axis.
The total charge on the rod is \(-Q\). The electric field due to a negative charge points towards the charge. Since the arc is in the positive x-region, the net field at the origin O will point towards the arc, i.e., in the positive x-direction (\( \hat{i} \)).
2. Linear Charge Density (\( \lambda \)):
The charge is distributed over the length of the arc.
Arc length \( L = R \times (\text{angle in radians}) \).
Total angle = \( 120^{\circ} = 120 \times \frac{\pi}{180} = \frac{2\pi}{3} \) radians.
\( L = R \frac{2\pi}{3} \).
Linear charge density \( \lambda = \frac{\text{Total Charge}}{\text{Length}} = \frac{-Q}{2\pi R/3} = -\frac{3Q}{2\pi R} \).
For calculating the magnitude of the field, we use the magnitude of the charge density, \( |\lambda| = \frac{3Q}{2\pi R} \).
3. Calculate Electric Field Magnitude:
Using the formula for the field of an arc:
\[ E = \frac{2k|\lambda|}{R} \sin(\alpha) \] Substitute \( k = \frac{1}{4\pi\varepsilon_0} \), \( |\lambda| = \frac{3Q}{2\pi R} \), and \( \alpha = 60^{\circ} \).
\[ E = \frac{2 \left( \frac{1}{4\pi\varepsilon_0} \right) \left( \frac{3Q}{2\pi R} \right)}{R} \sin(60^{\circ}) \] \[ E = \frac{2 \cdot 3Q}{4\pi\varepsilon_0 \cdot 2\pi R^2} \sin(60^{\circ}) \] \[ E = \frac{6Q}{8\pi^2\varepsilon_0 R^2} \left( \frac{\sqrt{3}}{2} \right) \] \[ E = \frac{3\sqrt{3}Q}{8\pi^2\varepsilon_0 R^2} \] 4. Determine the Direction:
As established from symmetry and the negative sign of the charge, the field vector at the origin points towards the arc, along the positive x-axis.
So, \( \vec{E} = E \hat{i} \).
\[ \vec{E} = \frac{3\sqrt{3} Q}{8 \pi^2 \varepsilon_0 R^2} (\hat{i}) \] Step 4: Final Answer:
The electric field at the centre of curvature O is \( \frac{3\sqrt{3} Q}{8 \pi^2 \varepsilon_0 R^2} (\hat{i}) \).