Let `V_0` be the potential at the origin in an electric field `vecE = 4hati + 5hatj`. The potential at the point (x,y) is

Options

A. `V_0 - 4x - 5y`
B. `V_0 + 4x + 5y`
C. `4x + 5y - V_0`
D. `-4x - 5y`

Explanation

To find the potential at the point (x, y) in the given electric field \(\vec{E} = 4\hat{i} + 5\hat{j}\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electric Field and Potential Relationship**:
The relationship between electric field \(\vec{E}\) and electric potential \(V\) is given by:
\[ \vec{E} = -\nabla V \]
or in differential form:
\[ dV = -\vec{E} \cdot d\vec{r} \]

2. **Identifying the Differential Displacement**:
The differential displacement vector \(d\vec{r}\) from the origin to the point (x, y) can be expressed as:
\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]

3. **Substituting the Electric Field**:
Substitute \(\vec{E}\) into the equation:
\[ dV = -\vec{E} \cdot d\vec{r} = - (4\hat{i} + 5\hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]

4. **Calculating the Dot Product**:
The dot product simplifies to:
\[ dV = - (4dx + 5dy) \]

5. **Integrating to Find the Potential**:
To find the potential difference from the origin (where the potential is \(V_0\)) to the point (x, y), we integrate:
\[ V(x, y) - V_0 = -\int_0^x 4 \, dx - \int_0^y 5 \, dy \]

6. **Performing the Integrals**:
- The integral of \(4 \, dx\) from \(0\) to \(x\) is:
\[ -\int_0^x 4 \, dx = -4x \]
- The integral of \(5 \, dy\) from \(0\) to \(y\) is:
\[ -\int_0^y 5 \, dy = -5y \]

7. **Combining the Results**:
Combining these results gives:
\[ V(x, y) - V_0 = -4x - 5y \]
Therefore, we can express \(V(x, y)\) as:
\[ V(x, y) = V_0 - 4x - 5y \]

### Final Answer:
The potential at the point (x, y) is:
\[ V(x, y) = V_0 - 4x - 5y \]