Let `V_(0)` be the potential at the origin in an electric field `vecE=E_(x)hati+E_(y)hatj`. The potential at the point `(x,y)` is

Options

A. `V_(@)+xE_(x)+yE_(y)`
B. `V_(@)+xE_(x)-yE_(y)`
C. `V_(@)-xE_(x)-yE_(y)`
D. `sqrt((x^(2)+y^(2)))sqrt((E_(x)^(2)+E_(y)^(2)))-V_(@)`

Answer

To find the potential at the point \((x, y)\) in an electric field given by \(\vec{E} = E_x \hat{i} + E_y \hat{j}\), we can follow these steps:

### Step-by-Step Solution

1. **Understand the Relationship Between Electric Field and Potential**:
The electric potential \(V\) at a point in an electric field is related to the electric field \(\vec{E}\) by the equation:
\[ V_P - V_0 = -\int_{0}^{P} \vec{E} \cdot d\vec{r} \]
where \(V_P\) is the potential at point \(P\), \(V_0\) is the potential at the origin, and \(d\vec{r}\) is the differential displacement vector.

2. **Define the Position Vector**:
The position vector of point \(P\) can be expressed as:
\[ \vec{r} = x \hat{i} + y \hat{j} \]

3. **Calculate the Change in Potential**:
The change in potential from the origin to point \(P\) can be expressed as:
\[ V_P - V_0 = -\int_{0}^{P} \vec{E} \cdot d\vec{r} \]
Here, we can express \(\vec{E}\) as:
\[ \vec{E} = E_x \hat{i} + E_y \hat{j} \]

4. **Evaluate the Dot Product**:
The differential displacement \(d\vec{r}\) can be expressed as:
\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]
Therefore, the dot product \(\vec{E} \cdot d\vec{r}\) becomes:
\[ \vec{E} \cdot d\vec{r} = (E_x \hat{i} + E_y \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = E_x dx + E_y dy \]

5. **Integrate to Find the Potential**:
Now we can integrate:
\[ V_P - V_0 = -\int_{0}^{P} (E_x dx + E_y dy) \]
This can be separated into two integrals:
\[ V_P - V_0 = -\left( \int_{0}^{x} E_x dx + \int_{0}^{y} E_y dy \right) \]

6. **Evaluate the Integrals**:
Assuming \(E_x\) and \(E_y\) are constant (or can be treated as such over the path), we can evaluate the integrals:
\[ V_P - V_0 = -\left( E_x x + E_y y \right) \]

7. **Express the Potential at Point \(P\)**:
Rearranging gives us the potential at point \(P\):
\[ V_P = V_0 - (E_x x + E_y y) \]

### Final Answer
Thus, the potential at the point \((x, y)\) is:
\[ V_P = V_0 - (E_x x + E_y y) \]