The correct relation between electric intensity E and electric potential V is

Options

A. `E = -(dV)/(dr)`
B. `E = (dV)/(dr)`
C. `V = -(dE)/(dr)`
D. `V = (dE)/(dr)`

Explanation

To find the correct relation between electric intensity \( E \) and electric potential \( V \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Electric Field and Potential**:
- Electric field intensity \( E \) is defined as the force per unit charge experienced by a small positive test charge placed in an electric field.
- Electric potential \( V \) at a point in space is the work done in bringing a unit positive charge from infinity to that point against the electric field.

2. **Relation Between Electric Field and Potential**:
- The electric field \( E \) is related to the electric potential \( V \) through the concept of the gradient. The electric field is defined as the negative gradient of the electric potential.
- Mathematically, this is expressed as:
\[ E = -\frac{dV}{dr} \]
- Here, \( \frac{dV}{dr} \) represents the rate of change of electric potential with respect to distance \( r \).

3. **Direction of Electric Field**:
- The negative sign indicates that the electric field points in the direction of decreasing potential. This means that if you move in the direction of the electric field, the potential decreases.

4. **Conclusion**:
- The correct relation between electric intensity \( E \) and electric potential \( V \) is:
\[ E = -\frac{dV}{dr} \]

### Final Answer:
The correct relation between electric intensity \( E \) and electric potential \( V \) is:
\[ E = -\frac{dV}{dr} \]