A wire with a linear charge density lambda forms a quarter of a cjrclenof radius R and is placed asa shown in thefigure with its centre of curvature at the origin O the electric field at O will be

To determine the electric field at the center of curvature (point O) due to a wire with a linear charge density \( \lambda \) that forms a quarter of a circle, we can break down the problem step by step. The key here is to understand how the electric field contributions from different segments of the wire combine at point O.

Understanding the Setup

Imagine a quarter-circle wire with radius \( R \) lying in the first quadrant of the Cartesian plane, with its center at the origin O. The wire carries a uniform linear charge density \( \lambda \). The goal is to find the resultant electric field at point O due to this charged wire.

Electric Field Contribution from a Small Segment

Consider a small segment of the wire, \( d\theta \), at an angle \( \theta \) from the positive x-axis. The length of this segment can be expressed as \( R \, d\theta \). The charge \( dq \) on this segment is given by:

• \( dq = \lambda \cdot R \, d\theta \)

The electric field \( d\mathbf{E} \) produced by this small charge segment at point O can be calculated using Coulomb's law. The magnitude of the electric field due to \( dq \) at point O is:

• \( dE = \frac{k \cdot dq}{R^2} = \frac{k \cdot \lambda \cdot R \, d\theta}{R^2} = \frac{k \cdot \lambda \, d\theta}{R} \)

Here, \( k \) is Coulomb's constant.

Direction of the Electric Field

The direction of \( d\mathbf{E} \) is radially outward from the charge segment. Since we are dealing with a quarter-circle, the electric field components can be resolved into x and y components:

• \( dE_x = dE \cos(\theta) \)
• \( dE_y = dE \sin(\theta) \)

Integrating Over the Quarter-Circle

To find the total electric field at point O, we need to integrate these components over the angle from \( 0 \) to \( \frac{\pi}{2} \):

• \( E_x = \int_0^{\frac{\pi}{2}} dE_x = \int_0^{\frac{\pi}{2}} \frac{k \cdot \lambda \, d\theta}{R} \cos(\theta) \)
• \( E_y = \int_0^{\frac{\pi}{2}} dE_y = \int_0^{\frac{\pi}{2}} \frac{k \cdot \lambda \, d\theta}{R} \sin(\theta) \)

Calculating the Integrals

Now, let's compute these integrals:

• For \( E_x \):

  \( E_x = \frac{k \cdot \lambda}{R} \int_0^{\frac{\pi}{2}} \cos(\theta) \, d\theta = \frac{k \cdot \lambda}{R} [\sin(\theta)]_0^{\frac{\pi}{2}} = \frac{k \cdot \lambda}{R} (1 - 0) = \frac{k \cdot \lambda}{R} \)

• For \( E_y \):

  \( E_y = \frac{k \cdot \lambda}{R} \int_0^{\frac{\pi}{2}} \sin(\theta) \, d\theta = \frac{k \cdot \lambda}{R} [-\cos(\theta)]_0^{\frac{\pi}{2}} = \frac{k \cdot \lambda}{R} (0 - (-1)) = \frac{k \cdot \lambda}{R} \)

Resultant Electric Field

The total electric field at point O is the vector sum of \( E_x \) and \( E_y \):

• \( \mathbf{E} = E_x \hat{i} + E_y \hat{j} = \frac{k \cdot \lambda}{R} \hat{i} + \frac{k \cdot \lambda}{R} \hat{j} \)

This can be expressed in magnitude and direction. The magnitude of the resultant electric field is:

• \( E = \sqrt{E_x^2 + E_y^2} = \sqrt{ \left( \frac{k \cdot \lambda}{R} \right)^2 + \left( \frac{k \cdot \lambda}{R} \right)^2 } = \sqrt{2} \cdot \frac{k \cdot \lambda}{R} \)

Final Expression

Thus, the electric field at point O due to the quarter-circle wire is:

• \( \mathbf{E} = \frac{k \cdot \lambda}{R} \sqrt{2} \hat{i} + \frac{k \cdot \lambda}{R} \sqrt{2} \hat{j} \)

This result shows that the electric field at the center of curvature is directed equally in the x and y directions, reflecting the symmetry of the charge distribution.