A point charge q is placed at the origin. Calculate the electric flux through the open hemispherical surface `(x-a)^2+y^2+z^2=a^2,xgea`

Explanation

To calculate the electric flux through the open hemispherical surface defined by the equation \((x-a)^2 + y^2 + z^2 = a^2\) for \(x \geq a\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- A point charge \(q\) is placed at the origin \((0, 0, 0)\).
- The open hemispherical surface is centered at \((a, 0, 0)\) with a radius \(a\) and extends in the positive \(x\) direction.

2. **Applying Gauss's Law**:
- According to Gauss's Law, the total electric flux \(\Phi\) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Here, \(Q_{\text{enc}} = q\) (the charge at the origin) and \(\epsilon_0\) is the permittivity of free space.

3. **Calculating the Total Flux**:
- The total flux through a closed surface surrounding the charge \(q\) (like a full sphere) is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

4. **Finding the Flux through the Open Hemisphere**:
- The open hemisphere only covers half of the total solid angle of a sphere. The solid angle for a full sphere is \(4\pi\) steradians, so the solid angle for a hemisphere is \(2\pi\) steradians.
- Therefore, the flux through the open hemisphere is half of the total flux:
\[ \Phi_{\text{hemisphere}} = \frac{1}{2} \cdot \frac{q}{\epsilon_0} = \frac{q}{2\epsilon_0} \]

5. **Considering the Orientation**:
- Since the hemisphere is open and oriented towards the positive \(x\) direction, the flux through the flat circular base of the hemisphere does not contribute to the total flux calculation, as it does not enclose any charge.

### Final Answer:
The electric flux through the open hemispherical surface is:
\[ \Phi = \frac{q}{2\epsilon_0} \]