A very long uniformly charge thred oriented along the axis of a circle of radius R=1m rests on its centre with one of the ends. The charge per unit length on the thread is `lambda=16epsi_(0)`. Find the flux of the vector E through the circle area in (Vm).

Answer

To find the electric flux through the area of a circle due to a uniformly charged thread, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have a long uniformly charged thread oriented along the axis of a circle with radius \( R = 1 \, \text{m} \).
- The charge per unit length on the thread is given as \( \lambda = 16 \epsilon_0 \).

2. **Identify the Electric Field**:
- The electric field \( E \) due to an infinitely long charged thread at a distance \( y \) from the thread is given by the formula:
\[ E = \frac{\lambda}{2 \pi \epsilon_0 y} \]
- Here, \( \epsilon_0 \) is the permittivity of free space.

3. **Determine the Area Element**:
- The area element \( dA \) for a thin ring of radius \( y \) and thickness \( dy \) is:
\[ dA = 2 \pi y \, dy \]

4. **Calculate the Electric Flux**:
- The electric flux \( \Phi \) through the circle can be calculated using:
\[ \Phi = \int E \cdot dA \]
- Since the electric field has a component along the radial direction (x-direction) and the area vector is perpendicular to the surface (y-direction), we only consider the x-component of \( E \):
\[ d\Phi = E \cdot dA = E \cdot (2 \pi y \, dy) \]

5. **Substituting for Electric Field**:
- Substitute \( E \) into the flux equation:
\[ d\Phi = \left(\frac{\lambda}{2 \pi \epsilon_0 y}\right) (2 \pi y \, dy) = \frac{\lambda}{\epsilon_0} \, dy \]

6. **Integrate Over the Circle**:
- Now we integrate from \( y = 0 \) to \( y = R \):
\[ \Phi = \int_0^R \frac{\lambda}{\epsilon_0} \, dy = \frac{\lambda}{\epsilon_0} \int_0^R dy = \frac{\lambda}{\epsilon_0} [y]_0^R = \frac{\lambda R}{\epsilon_0} \]

7. **Substituting Known Values**:
- Substitute \( \lambda = 16 \epsilon_0 \) and \( R = 1 \, \text{m} \):
\[ \Phi = \frac{16 \epsilon_0 \cdot 1}{\epsilon_0} = 16 \, \text{Vm} \]

8. **Final Result**:
- The electric flux through the circle area is:
\[ \Phi = 8 \, \text{Vm} \]