A positive charge Q is uniformly distributed across a rod of length L. The rod is then bent in the form of a semicircle. What will be the electric field intensity at the centre of curvature of the semicircle thus formed?

Explanation

To find the electric field intensity at the center of curvature of a semicircular rod with a uniformly distributed positive charge \( Q \), we can follow these steps:

### Step 1: Understand the Geometry
The rod of length \( L \) is bent into a semicircle. The radius \( R \) of the semicircle can be determined using the relationship between the length of the semicircle and its radius:
\[ L = \pi R \implies R = \frac{L}{\pi} \]

...