A charge 'Q' is placed at the centre of a hemispherical surface of radius 'R'. The flux of electric field due to charge 'Q' through the surface of hemisphere is

Options

A. `Q//4epsilon_(0)`
B. `Q//4piepsilon_(0)`
C. `Q//2epsilon_(0)`
D. `Q//2piepsilon_(0)`

Answer

To solve the problem of finding the electric flux through a hemispherical surface due to a charge \( Q \) placed at its center, we can follow these steps:

### Step 1: Understand the concept of electric flux
Electric flux (\( \Phi \)) through a surface is defined as the total number of electric field lines passing through that surface. Mathematically, it is given by:
\[ \Phi = \int \vec{E} \cdot d\vec{A} \]
where \( \vec{E} \) is the electric field and \( d\vec{A} \) is the differential area vector.

### Step 2: Use Gauss's Law
According to Gauss's Law, the total electric flux (\( \Phi \)) through a closed surface is proportional to the charge enclosed within that surface:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the charge enclosed and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Consider the entire spherical surface
If we consider a complete sphere of radius \( R \) with charge \( Q \) at its center, the total electric flux through the entire sphere would be:
\[ \Phi_{\text{sphere}} = \frac{Q}{\epsilon_0} \]

### Step 4: Relate the flux through the hemisphere to the sphere
Since the hemispherical surface is half of the entire sphere, the flux through the hemispherical surface would be half of the total flux through the sphere:
\[ \Phi_{\text{hemisphere}} = \frac{1}{2} \Phi_{\text{sphere}} = \frac{1}{2} \left(\frac{Q}{\epsilon_0}\right) \]
Thus, we have:
\[ \Phi_{\text{hemisphere}} = \frac{Q}{2\epsilon_0} \]

### Conclusion
The electric flux through the hemispherical surface due to the charge \( Q \) placed at its center is:
\[ \Phi = \frac{Q}{2\epsilon_0} \]