If `vec(E)=2x^(2) hat(i)-3y^(2) hat(j)`, then V (x, y, z)

Answer

To find the electric potential \( V(x, y, z) \) from the given electric field \( \vec{E} = 2x^2 \hat{i} - 3y^2 \hat{j} \), 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:
\[ \vec{E} = -\nabla V \]
This means that the electric field is the negative gradient of the potential.

### Step 2: Write the expression for the differential displacement vector
The differential displacement vector \( d\vec{r} \) can be expressed as:
\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]

### Step 3: Set up the integral for potential
The potential difference \( V \) can be expressed as:
\[ V = -\int \vec{E} \cdot d\vec{r} \]
Substituting the expression for \( \vec{E} \) and \( d\vec{r} \):
\[ V = -\int (2x^2 \hat{i} - 3y^2 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]

### Step 4: Calculate the dot product
Calculating the dot product:
\[ \vec{E} \cdot d\vec{r} = 2x^2 dx - 3y^2 dy \]
Thus, we can rewrite the integral:
\[ V = -\int (2x^2 dx - 3y^2 dy) \]

### Step 5: Separate the integrals
We can separate the integrals:
\[ V = -\left( \int 2x^2 dx - \int 3y^2 dy \right) \]

### Step 6: Perform the integrations
Now, we perform the integrations:
1. For the first integral:
\[ \int 2x^2 dx = \frac{2}{3} x^3 \]
2. For the second integral:
\[ \int 3y^2 dy = y^3 \]

### Step 7: Combine the results
Combining the results from the integrations:
\[ V = -\left( \frac{2}{3} x^3 - y^3 \right) + C \]
Where \( C \) is the constant of integration.

### Step 8: Write the final expression for potential
Thus, the potential \( V(x, y, z) \) is:
\[ V(x, y, z) = -\frac{2}{3} x^3 + y^3 + C \]

### Final Result
\[ V(x, y, z) = -\frac{2}{3} x^3 + y^3 + C \]

---