An electric field is expressed as `vec E = 2 hat i + 3 hat j`. Find the potential difference `(V_A - V_B)` between two points `A` and `B` whose position vectors are given by `r_A = hat i + 2 hat j and r_B = 2 hat i + hat j + 3 hat k`.

Options

A. `-1 V`
B. 1 V
C. 2 V
D. 3 V

Explanation

To find the potential difference \( V_A - V_B \) between two points \( A \) and \( B \) in an electric field given by \( \vec{E} = 2 \hat{i} + 3 \hat{j} \), we can use the formula:

\[ V_A - V_B = -\int_{A}^{B} \vec{E} \cdot d\vec{l} \]

### Step 1: Identify the position vectors
The position vectors are given as:
- \( \vec{r}_A = \hat{i} + 2 \hat{j} \)
- \( \vec{r}_B = 2 \hat{i} + \hat{j} + 3 \hat{k} \)

### Step 2: Determine the differential path element
The differential path element \( d\vec{l} \) can be expressed in terms of its components as:
\[ d\vec{l} = dx \hat{i} + dy \hat{j} + dz \hat{k} \]

### Step 3: Set up the integral
Since the electric field \( \vec{E} \) has no \( k \) component, we can ignore \( dz \) in our calculations. The limits of integration will be from point \( A \) to point \( B \).

### Step 4: Calculate the integral
We need to evaluate the integral:
\[ V_A - V_B = -\int_{A}^{B} (2 \hat{i} + 3 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]
This simplifies to:
\[ V_A - V_B = -\int_{A}^{B} (2dx + 3dy) \]

### Step 5: Determine the limits of integration
From \( \vec{r}_A \) to \( \vec{r}_B \):
- For \( x \): from \( 1 \) to \( 2 \)
- For \( y \): from \( 2 \) to \( 1 \)

### Step 6: Evaluate the integral
Now we can evaluate the integral:
\[ V_A - V_B = -\left( \int_{1}^{2} 2 \, dx + \int_{2}^{1} 3 \, dy \right) \]

Calculating the first integral:
\[ \int_{1}^{2} 2 \, dx = 2[x]_{1}^{2} = 2(2 - 1) = 2 \]

Calculating the second integral:
\[ \int_{2}^{1} 3 \, dy = 3[y]_{2}^{1} = 3(1 - 2) = -3 \]

### Step 7: Combine the results
Now substituting back into the equation:
\[ V_A - V_B = -\left( 2 - 3 \right) = -(-1) = 1 \]

Thus, the potential difference \( V_A - V_B \) is:
\[ \boxed{1 \, \text{volt}} \]