If q is the charge per unit area on the surface of a conductor, then the electric field intensity at a point on the surface is

Options

A. `((q)/(epsilon_(0)))` normal to surface
B. `((q)/(2epsilon_(0)))` normal to surface
C. `((q)/(epsilon_(0)))` tangential to surface
D. `((q)/(2epsilon_(0)))` tangential to surface

Answer

To determine the electric field intensity at a point on the surface of a conductor when given the charge per unit area \( q \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Charge Distribution**:
- A conductor in electrostatic equilibrium has charges distributed uniformly on its surface. The charge per unit area is denoted as \( q \).

2. **Identify the Electric Field Direction**:
- The electric field lines due to positive charges on the surface of the conductor point outward, perpendicular to the surface.

3. **Apply Gauss's Law**:
- According to Gauss's Law, the electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q_{\text{enc}} \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]

4. **Consider a Gaussian Surface**:
- To find the electric field at the surface, consider a small area \( dS \) on the surface of the conductor. The charge \( dQ \) on this area can be expressed as:
\[ dQ = q \cdot dS \]

5. **Calculate the Electric Flux**:
- The electric flux through the area \( dS \) is given by:
\[ \Phi = E \cdot dS \]
- Here, \( E \) is the electric field intensity at the surface.

6. **Relate Charge and Electric Field**:
- From Gauss's Law, substituting \( dQ \) into the equation gives:
\[ E \cdot dS = \frac{q \cdot dS}{\epsilon_0} \]

7. **Simplify the Equation**:
- Since \( dS \) is common on both sides, we can cancel it out (assuming \( dS \neq 0 \)):
\[ E = \frac{q}{\epsilon_0} \]

8. **Conclusion**:
- The electric field intensity \( E \) at a point on the surface of the conductor is given by:
\[ E = \frac{q}{\epsilon_0} \]
- The direction of the electric field is normal (perpendicular) to the surface of the conductor.

### Final Answer:
The electric field intensity at a point on the surface of a conductor is \( E = \frac{q}{\epsilon_0} \), directed outward from the surface.

---