A thin half ring of radius `R = 20 cm` is uniformly charged with a total charge `q = 0.70 nC`.Find the magnitude of the electric field strength at the curvaite centre of this half-ring.

Explanation

To find the magnitude of the electric field strength at the curvature center of a uniformly charged half ring, we can follow these steps:

### Step 1: Convert the given values to standard units
- The radius \( R \) of the half ring is given as \( 20 \, \text{cm} \). Convert this to meters:
\[ R = 20 \, \text{cm} = 0.20 \, \text{m} \]
- The total charge \( q \) on the half ring is \( 0.70 \, \text{nC} \). Convert this to coulombs:
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