The potential field of an electric field `vec(E)=(y hat(i)+x hat(j))` is

Options

A. `V=-xy+`constant
B. `V=-(x+y)+` constant
C. `V=-(x^(2)+y^(2))+` constant
D. V = constant

Answer

To find the potential field \( V \) corresponding to the given electric field \( \vec{E} = (y \hat{i} + x \hat{j}) \), we can follow these steps:

### Step 1: Understand the relationship between electric field and potential
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the equation:
\[ dV = -\vec{E} \cdot d\vec{r} \]
where \( d\vec{r} = dx \hat{i} + dy \hat{j} + dz \hat{k} \).

### Step 2: Substitute the electric field
Substituting the expression for \( \vec{E} \):
\[ dV = - (y \hat{i} + x \hat{j}) \cdot (dx \hat{i} + dy \hat{j} + dz \hat{k}) \]

### Step 3: Calculate the dot product
Calculating the dot product:
\[ dV = - (y \cdot dx + x \cdot dy + 0 \cdot dz) \]
This simplifies to:
\[ dV = - (y \, dx + x \, dy) \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ dV = -y \, dx - x \, dy \]

### Step 5: Integrate both sides
To find \( V \), we need to integrate:
\[ V = -\int (y \, dx + x \, dy) \]

### Step 6: Perform the integration
1. First, integrate with respect to \( x \):
\[ V = -\int y \, dx = -yx + f(y) \]
where \( f(y) \) is an arbitrary function of \( y \).

2. Now, differentiate \( V \) with respect to \( y \) to find \( f(y) \):
\[ \frac{\partial V}{\partial y} = -x + \frac{df(y)}{dy} \]
Setting this equal to \( -x \) (since \( \vec{E} = -\nabla V \)):
\[ -x + \frac{df(y)}{dy} = -x \]
This implies:
\[ \frac{df(y)}{dy} = 0 \implies f(y) = C \]
where \( C \) is a constant.

### Step 7: Write the final expression for potential
Thus, the potential \( V \) is:
\[ V = -xy + C \]

### Final Answer
The potential field corresponding to the electric field \( \vec{E} = (y \hat{i} + x \hat{j}) \) is:
\[ V = -xy + C \]