91Ó°ÊÓ

When a rod is curved into a semicircle, understanding the charge distribution is crucial. The entire length of the rod is bent into a semicircular shape, and to determine the radius of the semicircle, we need to know the length of the rod, which is given as \( L \).
Since the circumference of a semicircle is \( \pi R \), it follows that the radius \( R \) can be expressed as \( R = \frac{L}{\pi} \).
In this case, the charge \( Q \) is uniformly distributed over the length \( L \). This indicates that each segment of the rod has an equal amount of charge per unit length, denoted as \( \lambda = \frac{Q}{L} \). Understanding this distribution helps us calculate the electric field effectively when dealing with curved shapes.

Coulomb's Law is fundamental to understanding electric forces. It states that the force \( F \) between two point charges \( q_1 \) and \( q_2 \) is directly proportional to the product of their charges and inversely proportional to the square of the distance \( r \) between them. This is expressed mathematically as:\[F = k_e \frac{q_1 q_2}{r^2}\]Where \( k_e \) is Coulomb's constant, approximately equal to \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
In the context of a semicircular charge distribution, we are interested in how every small charge element, \( dq \), on the semicircle contributes to the electric field at the center.
By applying Coulomb's Law, we can find the electric field \( dE \) due to an infinitesimal charge element, \( dq \), as:\[dE = \frac{k_e \cdot dq}{R^2}\] Here, \( dq \) is calculated based on the linear charge density \( \lambda \) and the little arc segment \( ds \) it covers.

To find the total electric field at the center of curvature of the semicircle, integration is key. We take infinitesimal charge elements \( dq \), each contributing a small electric field \( dE \) at the center.
Due to symmetry, horizontal components of \( dE \) across the semicircle add up, while vertical components cancel each other. This simplifies the problem because you only need to integrate the horizontal components.

• Let \( dE_x = dE \cos\theta \), which focuses on the horizontal contribution.

In our problem, we integrate over the angle \( \theta \) from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\). The formula becomes:\[E_x = \int_{-\pi/2}^{\pi/2} \frac{k_e \cdot Q}{L \cdot R} \cos\theta \, d\theta\]Solving this integral yields the total electric field's magnitude at the center,\[E_x = \frac{2 \cdot k_e \cdot Q \cdot \pi}{L^2}\]This integrates the individual contributions into a single value, emphasizing the power of integration in electric field calculations.