If unifrom electric filed `vecE = E_(0)hati + 2E_(0) hatj`, where `E_(0)` is a constant, exists in a region of space and at `(0, 0)` the electric potential `V` is zero, then the potential at `(x_(0),0)` will be.

Options

A. Zero
B. `-E_(0)x_(0)`
C. `-2E_(0)x_(0)`
D. `-sqrt(5) E_(0)x_(0)`

Answer

To solve the problem, we need to find the electric potential at the point \((x_0, 0)\) given that the electric potential at the origin \((0, 0)\) is zero and the uniform electric field is \(\vec{E} = E_0 \hat{i} + 2E_0 \hat{j}\).

### Step-by-Step Solution:

1. **Understand the Electric Field**:
The electric field is given as \(\vec{E} = E_0 \hat{i} + 2E_0 \hat{j}\). This means:
- The electric field in the x-direction, \(E_x = E_0\).
- The electric field in the y-direction, \(E_y = 2E_0\).

2. **Use the Relation Between Electric Field and Potential**:
The relationship between electric field and electric potential is given by:
\[ E_x = -\frac{dV}{dx} \]
This implies:
\[ E_x \, dx = -dV \]

3. **Calculate the Change in Potential**:
We want to find the potential at the point \((x_0, 0)\). Since the potential at the origin \((0, 0)\) is \(V(0, 0) = 0\), we can express the change in potential as:
\[ dV = -E_x \, dx \]
Here, \(dx\) is the change in the x-coordinate from \(0\) to \(x_0\), so \(dx = x_0 - 0 = x_0\).

4. **Substitute the Values**:
Substitute \(E_x = E_0\) into the equation:
\[ dV = -E_0 \, dx \]
Therefore, we have:
\[ dV = -E_0 \, x_0 \]

5. **Find the Potential at \((x_0, 0)\)**:
The potential at point \((x_0, 0)\) can be expressed as:
\[ V(x_0, 0) = V(0, 0) + dV \]
Since \(V(0, 0) = 0\), we get:
\[ V(x_0, 0) = 0 - E_0 \, x_0 = -E_0 \, x_0 \]

### Final Answer:
The potential at the point \((x_0, 0)\) is:
\[ V(x_0, 0) = -E_0 x_0 \]