A unifrom electric field having a magnitude `E_(0)` and direction along the positive X-axis exists. If the protential `V` is zero at `x = 0`, then its value at `X = +x` will be

Options

A. `+xE_(0)`
B. `-xE_(0)`
C. `+x^(2)E_(0)`
D. `-x^(2)E_(0)`

Answer

To solve the problem step by step, we need to understand the relationship between electric field and electric potential.

### Step-by-Step Solution:

1. **Understanding the Electric Field and Potential Relationship**:
The electric field \( E \) is related to the electric potential \( V \) by the equation:
\[ E = -\frac{dV}{dx} \]
This means that the electric field is equal to the negative rate of change of potential with respect to position.

2. **Setting Up the Problem**:
We are given that the electric field \( E_0 \) is uniform and directed along the positive X-axis. Since the electric field is uniform, we can take \( E = E_0 \).

3. **Integrating to Find Potential**:
We need to find the potential \( V \) at a position \( x \) given that \( V = 0 \) at \( x = 0 \). We can rearrange the relationship:
\[ dV = -E dx \]
Substituting \( E = E_0 \):
\[ dV = -E_0 dx \]

4. **Integrating from 0 to x**:
We integrate both sides from \( 0 \) to \( V \) and from \( 0 \) to \( x \):
\[ \int_{0}^{V} dV = -E_0 \int_{0}^{x} dx \]
This gives us:
\[ V - 0 = -E_0 (x - 0) \]
Simplifying this, we find:
\[ V = -E_0 x \]

5. **Final Result**:
Therefore, the potential \( V \) at position \( x \) is:
\[ V = -E_0 x \]