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

### Step 1: Understand the relationship between electric field and electric potential
The electric field \( \vec{E} \) is related to the electric potential \( V \) by the equation:
\[ \vec{E} = -\nabla V \]
This means that the electric field is the negative gradient of the electric potential.

### Step 2: Write the expression for the electric field
Given:
\[ \vec{E} = y \hat{i} + x \hat{j} \]

### Step 3: Express the differential change in potential
From the relationship, we can express the differential change in potential \( dV \) as:
\[ dV = -\vec{E} \cdot d\vec{r} \]
where \( d\vec{r} = dx \hat{i} + dy \hat{j} \).

### Step 4: Calculate the dot product
Now, we calculate the dot product \( \vec{E} \cdot d\vec{r} \):
\[ \vec{E} \cdot d\vec{r} = (y \hat{i} + x \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = y \, dx + x \, dy \]
Thus, we have:
\[ dV = -(y \, dx + x \, dy) \]

### Step 5: Integrate to find the potential
Now, we need to integrate \( dV \):
\[ dV = -y \, dx - x \, dy \]
We can integrate this expression. We can integrate with respect to \( x \) and \( y \) separately.

1. Integrate \( -y \, dx \):
\[ \int -y \, dx = -yx + C(y) \]
where \( C(y) \) is a function of \( y \).

2. Integrate \( -x \, dy \):
\[ \int -x \, dy = -xy + C(x) \]
where \( C(x) \) is a function of \( x \).

Combining these, we have:
\[ V = -xy + C \]
where \( C \) is a constant of integration.

### Step 6: Final expression for potential
Thus, the electric potential \( V \) is given by:
\[ V = -xy + C \]

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

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