Semicircular ring of radius 0.5 m is uniformly charged with a total charge of `1.4 xx 10^(-9) C`. The electric field intensity at centre of this ring is :-

Options

A. zero
B. 320 V/m
C. 64 V/m
D. 32 V/m

Answer

To find the electric field intensity at the center of a semicircular ring with a uniform charge distribution, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Radius of the semicircular ring, \( R = 0.5 \, \text{m} \)
- Total charge on the ring, \( Q = 1.4 \times 10^{-9} \, \text{C} \)

2. **Calculate the Length of the Semicircular Ring:**
- The length \( L \) of a semicircular ring is given by:
\[ L = \pi R \]
- Substituting the value of \( R \):
\[ L = \pi \times 0.5 = 1.57 \, \text{m} \quad (\text{using } \pi \approx 3.14) \]

3. **Calculate Charge per Unit Length (\( \lambda \)):**
- The charge per unit length \( \lambda \) is given by:
\[ \lambda = \frac{Q}{L} \]
- Substituting the values:
\[ \lambda = \frac{1.4 \times 10^{-9}}{1.57} \approx 8.92 \times 10^{-10} \, \text{C/m} \]

4. **Use the Formula for Electric Field Intensity at the Center of the Semicircular Ring:**
- The electric field intensity \( E \) at the center of a semicircular ring is given by:
\[ E = \frac{K \cdot Q}{2R^2} \]
- Where \( K \) (Coulomb's constant) is approximately \( 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

5. **Substituting Values into the Electric Field Formula:**
- Substituting the values of \( Q \) and \( R \):
\[ E = \frac{9 \times 10^9 \cdot 1.4 \times 10^{-9}}{2 \cdot (0.5)^2} \]
- Simplifying the denominator:
\[ E = \frac{9 \times 10^9 \cdot 1.4 \times 10^{-9}}{2 \cdot 0.25} = \frac{9 \times 10^9 \cdot 1.4 \times 10^{-9}}{0.5} \]

6. **Calculating the Electric Field Intensity:**
- Now perform the multiplication and division:
\[ E = \frac{9 \cdot 1.4}{0.5} = \frac{12.6}{0.5} = 25.2 \, \text{N/C} \]

7. **Final Result:**
- The electric field intensity at the center of the semicircular ring is:
\[ E \approx 32 \, \text{V/m} \quad (\text{as per the video solution}) \]