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

Explanation

To solve the problem, we need to determine the electric potential \( V \) at a point \( x = +x \) in a uniform electric field \( E_0 \) that is directed along the positive x-axis, given that the potential \( V \) is zero at \( x = 0 \).

### Step-by-Step Solution:

1. **Understanding the Electric Field and Potential Relationship**:
The relationship between electric field \( E \) and electric potential \( V \) is given by the equation:
\[ E = -\frac{dV}{dx} \]
This indicates that the electric field is the negative gradient of the electric potential.

2. **Setting Up the Problem**:
We know that the electric field \( E_0 \) is uniform and directed along the positive x-axis. Therefore, we can express it as:
\[ E = E_0 \quad \text{(since it's in the positive x direction)} \]

3. **Integrating to Find Potential**:
To find the potential \( V \) at a point \( x \), we can integrate the electric field:
\[ V(x) - V(0) = -\int_{0}^{x} E \, dx \]
Since \( V(0) = 0 \) (given), we have:
\[ V(x) = -\int_{0}^{x} E_0 \, dx \]

4. **Performing the Integration**:
The integral of a constant \( E_0 \) over the interval from 0 to \( x \) is:
\[ V(x) = -E_0 \cdot (x - 0) = -E_0 \cdot x \]

5. **Final Expression for Potential**:
Therefore, the potential at \( x = +x \) is:
\[ V(x) = -E_0 x \]

### Conclusion:
The value of the potential \( V \) at \( x = +x \) is:
\[ V = -E_0 x \]