The electric flux coming out of the equi-potential surface is -

Options

A. perpendicular to the surface
B. parallel to the surface
C. in all directions
D. zero

Answer

To solve the question regarding the electric flux coming out of an equipotential surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Equipotential Surfaces**:
- An equipotential surface is defined as a surface where the electric potential is constant. This means that no work is done when moving a charge along this surface.

2. **Work Done and Electric Field**:
- The work done \( W \) in moving a charge \( q \) in an electric field \( E \) is given by the formula:
\[ W = F \cdot ds = qE \cdot ds \cdot \cos(\theta) \]
- For the work done to be zero, \( \cos(\theta) \) must be zero, which occurs when \( \theta = 90^\circ \). This indicates that the force \( F \) (due to the electric field) and the displacement \( ds \) are perpendicular to each other.

3. **Direction of Electric Field**:
- The electric field lines are always directed from regions of higher potential to lower potential. Therefore, the electric field is perpendicular to the equipotential surfaces.

4. **Electric Flux**:
- Electric flux \( \Phi_E \) through a surface is defined as:
\[ \Phi_E = E \cdot A \cdot \cos(\phi) \]
- Here, \( E \) is the electric field strength, \( A \) is the area of the surface, and \( \phi \) is the angle between the electric field and the normal to the surface.

5. **Applying to Equipotential Surfaces**:
- Since the equipotential surface is perpendicular to the electric field, the angle \( \phi \) is \( 90^\circ \). Therefore, \( \cos(90^\circ) = 0 \).

6. **Conclusion**:
- As a result, the electric flux \( \Phi_E \) through the equipotential surface is:
\[ \Phi_E = E \cdot A \cdot \cos(90^\circ) = 0 \]
- Thus, the electric flux coming out of the equipotential surface is **zero**.

### Final Answer:
The electric flux coming out of the equipotential surface is **zero**.