An electric field of `20 N//C` exists along the x-axis in space. Calculate the potential difference `V_B - V_A` where the points A and B are given by
a. `A=(0,0), B=(4m, 2m)`
b. `A=(4m,2m),B=(6m,5m)`

Explanation

To solve the problem of calculating the potential difference \( V_B - V_A \) in an electric field of \( 20 \, \text{N/C} \) along the x-axis for the given points, we will follow these steps:

### Step-by-Step Solution

**Step 1: Understand the relationship between electric field and potential difference.**
The potential difference \( V_B - V_A \) between two points in an electric field can be calculated using the formula:
\[ V_B - V_A = - \int_A^B \mathbf{E} \cdot d\mathbf{r} \]
where \( \mathbf{E} \) is the electric field vector and \( d\mathbf{r} \) is the differential displacement vector.

**Step 2: Define the electric field vector.**
Given that the electric field \( \mathbf{E} \) is \( 20 \, \text{N/C} \) along the x-axis, we can express it as:
\[ \mathbf{E} = 20 \, \hat{i} \, \text{N/C} \]

**Step 3: Identify the points for part (a).**
For part (a), the points are:
- \( A = (0, 0) \)
- \( B = (4 \, \text{m}, 2 \, \text{m}) \)

**Step 4: Calculate the potential difference \( V_B - V_A \) for part (a).**
Since the electric field is only in the x-direction, the y-component does not contribute to the potential difference. Thus, we only consider the x-coordinates:
\[ V_B - V_A = - \int_{0}^{4} \mathbf{E} \cdot d\mathbf{r} \]
Here, \( d\mathbf{r} = dx \, \hat{i} + dy \, \hat{j} \). The dot product \( \mathbf{E} \cdot d\mathbf{r} = 20 \, dx \) since \( dy \) does not contribute.

Now, we can evaluate the integral:
\[ V_B - V_A = - \int_{0}^{4} 20 \, dx = -20 \left[ x \right]_{0}^{4} = -20 (4 - 0) = -80 \, \text{V} \]

**Step 5: Identify the points for part (b).**
For part (b), the points are:
- \( A = (4 \, \text{m}, 2 \, \text{m}) \)
- \( B = (6 \, \text{m}, 5 \, \text{m}) \)

**Step 6: Calculate the potential difference \( V_B - V_A \) for part (b).**
Again, we only consider the x-coordinates:
\[ V_B - V_A = - \int_{4}^{6} \mathbf{E} \cdot d\mathbf{r} \]
The dot product remains the same:
\[ \mathbf{E} \cdot d\mathbf{r} = 20 \, dx \]
Now we evaluate the integral:
\[ V_B - V_A = - \int_{4}^{6} 20 \, dx = -20 \left[ x \right]_{4}^{6} = -20 (6 - 4) = -20 \times 2 = -40 \, \text{V} \]

### Final Answers
- For part (a): \( V_B - V_A = -80 \, \text{V} \)
- For part (b): \( V_B - V_A = -40 \, \text{V} \)