If uniform electric field ` vec E = E_ 0 hati + 2 E _ 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 , 2x _ 0 ) ` will be -

Options

A. zero
B. ` - 5E _ 0 x _ 0 `
C. ` - 3 E _ 0 x _ 0 `
D. `-sqrt5 E _ 0 x _ 0 `

Explanation

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

### Step-by-Step Solution:

1. **Understand the Electric Field and Potential Relation**:
The potential difference \( \Delta V \) between two points in an electric field is given by:
\[ \Delta V = V(B) - V(A) = -\int_A^B \vec{E} \cdot d\vec{r} \]
where \( A \) is the starting point and \( B \) is the ending point.

2. **Identify the Points**:
Let point \( A \) be \( (0, 0) \) and point \( B \) be \( (x_0, 2x_0) \). The potential at point \( A \) is given as \( V(0, 0) = 0 \).

3. **Set Up the Integral**:
The differential displacement vector \( d\vec{r} \) can be expressed as:
\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]
The electric field \( \vec{E} \) is given as:
\[ \vec{E} = E_0 \hat{i} + 2E_0 \hat{j} \]

4. **Calculate the Dot Product**:
The dot product \( \vec{E} \cdot d\vec{r} \) is:
\[ \vec{E} \cdot d\vec{r} = (E_0 \hat{i} + 2E_0 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = E_0 dx + 2E_0 dy \]

5. **Set Up the Integral for Potential**:
The potential at point \( B \) can be calculated as:
\[ V(x_0, 2x_0) = V(0, 0) - \int_{(0,0)}^{(x_0, 2x_0)} \vec{E} \cdot d\vec{r} \]
Since \( V(0, 0) = 0 \), we have:
\[ V(x_0, 2x_0) = -\int_{0}^{x_0} E_0 \, dx - \int_{0}^{2x_0} 2E_0 \, dy \]

6. **Evaluate the Integrals**:
- For the first integral:
\[ -\int_{0}^{x_0} E_0 \, dx = -E_0 [x]_0^{x_0} = -E_0 x_0 \]
- For the second integral:
\[ -\int_{0}^{2x_0} 2E_0 \, dy = -2E_0 [y]_0^{2x_0} = -2E_0 (2x_0) = -4E_0 x_0 \]

7. **Combine the Results**:
Now, adding both contributions:
\[ V(x_0, 2x_0) = -E_0 x_0 - 4E_0 x_0 = -5E_0 x_0 \]

### Final Result:
Thus, the potential at the point \( (x_0, 2x_0) \) is:
\[ V(x_0, 2x_0) = -5E_0 x_0 \]