Potential difference between two points `(V_(A) - V_(B))` in an electric field `E = (2 hat(i) - 4 hat(j)) N//C`, where `A = (0, 0)` and `B = (3m, 4m)` is

Options

A. `10 V`
B. `-10 V`
C. `16 V`
D. `-16 V`

Answer

To find the potential difference between two points \( A \) and \( B \) in an electric field, we can use the formula:

\[ V_A - V_B = -\int_A^B \mathbf{E} \cdot d\mathbf{r} \]

### Step 1: Define the Electric Field and Points
Given:
- Electric field \( \mathbf{E} = 2 \hat{i} - 4 \hat{j} \, \text{N/C} \)
- Point \( A = (0, 0) \)
- Point \( B = (3 \, \text{m}, 4 \, \text{m}) \)

### Step 2: Set Up the Differential Displacement Vector
The differential displacement vector \( d\mathbf{r} \) can be expressed as:
\[ d\mathbf{r} = dx \hat{i} + dy \hat{j} \]

### Step 3: Substitute into the Integral
Now we substitute \( \mathbf{E} \) and \( d\mathbf{r} \) into the integral:
\[ V_A - V_B = -\int_A^B (2 \hat{i} - 4 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) \]

### Step 4: Calculate the Dot Product
Calculating the dot product:
\[ \mathbf{E} \cdot d\mathbf{r} = (2 \hat{i} - 4 \hat{j}) \cdot (dx \hat{i} + dy \hat{j}) = 2 \, dx - 4 \, dy \]

### Step 5: Set Up the Integral
Now we can set up the integral:
\[ V_A - V_B = -\int_A^B (2 \, dx - 4 \, dy) \]

### Step 6: Determine the Limits of Integration
For the path from point \( A(0, 0) \) to point \( B(3, 4) \), we can integrate \( x \) from \( 0 \) to \( 3 \) and \( y \) from \( 0 \) to \( 4 \).

### Step 7: Evaluate the Integral
Now we can evaluate the integral:
\[ V_A - V_B = -\left( \int_0^3 2 \, dx + \int_0^4 (-4) \, dy \right) \]

Calculating each integral:
1. \( \int_0^3 2 \, dx = 2x \big|_0^3 = 2(3) - 2(0) = 6 \)
2. \( \int_0^4 -4 \, dy = -4y \big|_0^4 = -4(4) - (-4)(0) = -16 \)

### Step 8: Combine the Results
Substituting back into the equation:
\[ V_A - V_B = -\left( 6 - 16 \right) = -(-10) = 10 \]

### Final Result
Thus, the potential difference \( V_A - V_B \) is:
\[ V_A - V_B = -10 \, \text{V} \]