The electric field strenth vector is given by E=`2.5x hat I + 1.5y hat`.The potential at point (2,2,1) is (considering potential at origin to be zero)

Options

A. `-2` V
B. `-4` V
C. `-6` V
D. `-8` V

Answer

To find the electric potential at the point (2, 2, 1) given the electric field vector \( \mathbf{E} = 2.5 \hat{i} + 1.5 \hat{j} \), we can use the relationship between electric potential \( V \) and electric field \( \mathbf{E} \):

\[ V = - \int \mathbf{E} \cdot d\mathbf{r} \]

### Step-by-Step Solution:

1. **Define the Electric Field and the Path of Integration**:
The electric field is given as \( \mathbf{E} = 2.5x \hat{i} + 1.5y \hat{j} \). We need to calculate the potential from the origin (0, 0, 0) to the point (2, 2, 1).

2. **Set Up the Integral**:
The potential difference can be calculated using the integral:
\[ V = - \int_{(0,0,0)}^{(2,2,1)} \mathbf{E} \cdot d\mathbf{r} \]
Here, \( d\mathbf{r} = dx \hat{i} + dy \hat{j} + dz \hat{k} \).

3. **Substituting the Electric Field**:
We can express the integral as:
\[ V = - \int_{0}^{2} (2.5x \, dx + 1.5y \, dy) \]
Since \( z \) does not appear in the electric field, we can ignore it for this calculation.

4. **Integrate with Respect to x and y**:
We will integrate \( x \) from 0 to 2 and \( y \) from 0 to 2:
\[ V = - \left( \int_{0}^{2} 2.5x \, dx + \int_{0}^{2} 1.5y \, dy \right) \]

5. **Calculate Each Integral**:
- For the first integral:
\[ \int_{0}^{2} 2.5x \, dx = 2.5 \left[ \frac{x^2}{2} \right]_{0}^{2} = 2.5 \left[ \frac{2^2}{2} - 0 \right] = 2.5 \cdot 2 = 5 \]
- For the second integral:
\[ \int_{0}^{2} 1.5y \, dy = 1.5 \left[ \frac{y^2}{2} \right]_{0}^{2} = 1.5 \left[ \frac{2^2}{2} - 0 \right] = 1.5 \cdot 2 = 3 \]

6. **Combine the Results**:
Now substituting back into the potential equation:
\[ V = - (5 + 3) = -8 \text{ volts} \]

Thus, the potential at the point (2, 2, 1) is \( -8 \) volts.