If the electric field in a region is given as `E=y^(2)hati+2xyhatj` and the potential is assumed `4` Volts at the origin, find the potential, in Volts, at the point `(2,1,9)`. All values are in SI units.

Explanation

To find the potential at the point (2, 1, 9) given the electric field \( \mathbf{E} = y^2 \hat{i} + 2xy \hat{j} \) and the potential at the origin (0, 0, 0) is 4 Volts, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electric Field and Potential Relationship:**
The relationship between electric field \( \mathbf{E} \) and electric potential \( V \) is given by:
\[ \mathbf{E} = -\nabla V \]
This implies that:
\[ E_x = -\frac{dV}{dx}, \quad E_y = -\frac{dV}{dy} \]

2. **Identifying Components of Electric Field:**
From the given electric field \( \mathbf{E} = y^2 \hat{i} + 2xy \hat{j} \), we can identify:
\[ E_x = y^2, \quad E_y = 2xy \]

3. **Finding the Potential Change:**
We can express the change in potential \( V \) as:
\[ V = V_0 - \int_{C} \mathbf{E} \cdot d\mathbf{r} \]
where \( V_0 \) is the potential at the origin (4 Volts), and \( C \) is the path from the origin to the point (2, 1, 9).

4. **Setting Up the Integrals:**
We will integrate along the x-axis and then along the y-axis. The total potential at point (2, 1, 9) can be calculated as:
\[ V(2, 1, 9) = V(0, 0, 0) - \left( \int_0^2 E_x \, dx + \int_0^1 E_y \, dy \right) \]
Substituting the expressions for \( E_x \) and \( E_y \):
\[ V(2, 1, 9) = 4 - \left( \int_0^2 y^2 \, dx + \int_0^1 2xy \, dy \right) \]

5. **Calculating the Integrals:**
- For \( \int_0^2 y^2 \, dx \) when \( y = 1 \):
\[ \int_0^2 1^2 \, dx = \int_0^2 1 \, dx = [x]_0^2 = 2 \]
- For \( \int_0^1 2xy \, dy \) when \( x = 2 \):
\[ \int_0^1 2(2)y \, dy = 4 \int_0^1 y \, dy = 4 \left[ \frac{y^2}{2} \right]_0^1 = 4 \cdot \frac{1}{2} = 2 \]

6. **Combining the Results:**
Now substituting back into the potential equation:
\[ V(2, 1, 9) = 4 - (2 + 2) = 4 - 4 = 0 \text{ Volts} \]

### Final Answer:
The potential at the point (2, 1, 9) is **0 Volts**.