Find the `V_(ab)` in an electric field `E=(2hati+3hatj+4hatk)N/C`, where `r_a=(hati-2hatj+hatk)m` and `r_b=(2hati+hatj-2hatk)m`

Explanation

To find the potential difference \( V_{ab} \) in the given electric field, we can follow these steps:

### Step 1: Identify the given quantities
We have:
- Electric field \( \mathbf{E} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \, \text{N/C} \)
- Position vector at point A \( \mathbf{r}_A = \hat{i} - 2 \hat{j} + \hat{k} \, \text{m} \)
- Position vector at point B \( \mathbf{r}_B = 2 \hat{i} + \hat{j} - 2 \hat{k} \, \text{m} \)

### Step 2: Calculate the change in position vector \( \mathbf{r}_B - \mathbf{r}_A \)
\[ \mathbf{r}_B - \mathbf{r}_A = (2 \hat{i} + \hat{j} - 2 \hat{k}) - (\hat{i} - 2 \hat{j} + \hat{k}) \]
Calculating this gives:
\[ \mathbf{r}_B - \mathbf{r}_A = (2 - 1) \hat{i} + (1 + 2) \hat{j} + (-2 - 1) \hat{k} = \hat{i} + 3 \hat{j} - 3 \hat{k} \]

### Step 3: Use the formula for potential difference
The potential difference \( V_{ab} \) is given by:
\[ V_{ab} = V_A - V_B = -\mathbf{E} \cdot (\mathbf{r}_B - \mathbf{r}_A) \]

### Step 4: Calculate the dot product \( \mathbf{E} \cdot (\mathbf{r}_B - \mathbf{r}_A) \)
Substituting the values:
\[ \mathbf{E} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \]
\[ \mathbf{r}_B - \mathbf{r}_A = \hat{i} + 3 \hat{j} - 3 \hat{k} \]
Now, calculate the dot product:
\[ \mathbf{E} \cdot (\mathbf{r}_B - \mathbf{r}_A) = (2 \hat{i} + 3 \hat{j} + 4 \hat{k}) \cdot (\hat{i} + 3 \hat{j} - 3 \hat{k}) \]
Calculating this gives:
\[ = 2 \cdot 1 + 3 \cdot 3 + 4 \cdot (-3) = 2 + 9 - 12 = -1 \]

### Step 5: Calculate \( V_{ab} \)
Now substituting back into the equation for \( V_{ab} \):
\[ V_{ab} = -(-1) = 1 \, \text{Volt} \]

### Final Answer
Thus, the potential difference \( V_{ab} \) is:
\[ V_{ab} = -1 \, \text{Volt} \]