The electric field in a certain region is given by `E=5 hat(i)-3hat(j) kv//m`. The potential difference `V_(B)-V_(A)` between points a and B having coordinates (4, 0, 3) m and (10, 3, 0) m respectively, is equal to

Options

A. 21 kV
B. `-21 kV`
C. `39 kV`
D. `-39 kV`

Answer

To find the potential difference \( V_B - V_A \) between points A and B in an electric field \( \mathbf{E} = 5 \hat{i} - 3 \hat{j} \) kV/m, we can use the formula:

\[ V_B - V_A = -\mathbf{E} \cdot \mathbf{R} \]

where \( \mathbf{R} \) is the displacement vector from point A to point B.

### Step 1: Identify the coordinates of points A and B
- Point A has coordinates \( (4, 0, 3) \) m.
- Point B has coordinates \( (10, 3, 0) \) m.

### Step 2: Determine the position vectors of points A and B
- Position vector of A: \( \mathbf{r}_A = 4 \hat{i} + 0 \hat{j} + 3 \hat{k} \)
- Position vector of B: \( \mathbf{r}_B = 10 \hat{i} + 3 \hat{j} + 0 \hat{k} \)

### Step 3: Calculate the displacement vector \( \mathbf{R} \)
\[ \mathbf{R} = \mathbf{r}_B - \mathbf{r}_A = (10 \hat{i} + 3 \hat{j} + 0 \hat{k}) - (4 \hat{i} + 0 \hat{j} + 3 \hat{k}) \]
\[ \mathbf{R} = (10 - 4) \hat{i} + (3 - 0) \hat{j} + (0 - 3) \hat{k} = 6 \hat{i} + 3 \hat{j} - 3 \hat{k} \]

### Step 4: Calculate the dot product \( \mathbf{E} \cdot \mathbf{R} \)
\[ \mathbf{E} = 5 \hat{i} - 3 \hat{j} \quad \text{(in kV/m)} \]
\[ \mathbf{E} \cdot \mathbf{R} = (5 \hat{i} - 3 \hat{j}) \cdot (6 \hat{i} + 3 \hat{j} - 3 \hat{k}) \]
Calculating the dot product:
\[ = 5 \cdot 6 + (-3) \cdot 3 + 0 \cdot (-3) = 30 - 9 + 0 = 21 \]

### Step 5: Calculate the potential difference
Using the formula for potential difference:
\[ V_B - V_A = -\mathbf{E} \cdot \mathbf{R} = -21 \text{ kV} \]

Thus, the potential difference \( V_B - V_A \) is:
\[ V_B - V_A = -21 \text{ kV} \]

### Final Answer
The potential difference \( V_B - V_A \) is \( -21 \text{ kV} \).

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