An electric field `vec(E)=B x hat(i)` exists in space, where `B = 20 V//m^(2)`. Taking the potential at (2m, 4m) to be zero, find the potential at the origin.

Answer

To find the potential at the origin given the electric field and the potential at a specific point, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Electric Field**:
The electric field is given as \( \vec{E} = B \hat{i} \) where \( B = 20 \, \text{V/m}^2 \). Therefore, we can express the electric field as:
\[ \vec{E} = 20 \hat{i} \, \text{V/m} \]

2. **Understand the Relationship Between Electric Field and Potential**:
The change in electric potential \( V \) between two points in an electric field is given by:
\[ V = - \int \vec{E} \cdot d\vec{r} \]
where \( d\vec{r} \) is the differential displacement vector.

3. **Set Up the Integral**:
We need to find the potential at the origin (0, 0) relative to the point (2 m, 4 m) where the potential is defined as zero. The displacement vector \( d\vec{r} \) can be expressed as:
\[ d\vec{r} = dx \hat{i} + dy \hat{j} \]
The path from (2, 4) to (0, 0) can be taken as a straight line along the x-axis first and then along the y-axis.

4. **Calculate the Potential Change**:
The potential at the origin can be calculated as:
\[ V(0, 0) - V(2, 4) = - \int_{(2, 4)}^{(0, 0)} \vec{E} \cdot d\vec{r} \]
Since \( V(2, 4) = 0 \), we have:
\[ V(0, 0) = - \int_{(2, 4)}^{(0, 0)} \vec{E} \cdot d\vec{r} \]

5. **Break Down the Integral**:
We can break the integral into two parts:
- From (2, 4) to (2, 0) (moving vertically down)
- From (2, 0) to (0, 0) (moving horizontally left)

For the vertical path (2, 4) to (2, 0):
\[ d\vec{r} = dy \hat{j} \quad \text{(where } dx = 0\text{)} \]
The electric field \( \vec{E} \) only has an x-component, so:
\[ \vec{E} \cdot d\vec{r} = 0 \quad \text{(no contribution)} \]

For the horizontal path (2, 0) to (0, 0):
\[ d\vec{r} = dx \hat{i} \quad \text{(where } dy = 0\text{)} \]
The integral becomes:
\[ V(0, 0) = - \int_{2}^{0} \vec{E} \cdot d\vec{r} = - \int_{2}^{0} 20 \, dx \]

6. **Evaluate the Integral**:
\[ V(0, 0) = -20 \int_{2}^{0} dx = -20 [x]_{2}^{0} = -20 (0 - 2) = 40 \, \text{V} \]

7. **Conclusion**:
Therefore, the potential at the origin is:
\[ V(0, 0) = 40 \, \text{V} \]