An electric field is given by `vec(E)=(yhat(i)+xhat(j))N/C`. Find the work done (in J) in moving a `1 C` charge from `vec(r)_(A)=(2hat(i)+2hat(j))` m to `vec(r)_(B)=(4hat(i)+hat(j))m`.

Options

A. `+4 J`
B. `-4 J`
C. `+8 J`
D. zero

Answer

To solve the problem of finding the work done in moving a 1 C charge from point A to point B in the given electric field, we can follow these steps:

### Step 1: Understand the Electric Field
The electric field is given by:
\[ \vec{E} = \hat{i}y + \hat{j}x \quad \text{(in N/C)} \]

### Step 2: Define the Positions
The initial position \( \vec{r}_A \) and final position \( \vec{r}_B \) are given as:
\[ \vec{r}_A = 2\hat{i} + 2\hat{j} \quad \text{(in m)} \]
\[ \vec{r}_B = 4\hat{i} + 1\hat{j} \quad \text{(in m)} \]

### Step 3: Calculate the Differential Displacement
The differential displacement \( d\vec{r} \) can be expressed as:
\[ d\vec{r} = dx\hat{i} + dy\hat{j} \]

### Step 4: Set Up the Work Done Formula
The work done \( W \) in moving a charge \( Q \) in an electric field is given by:
\[ W = -Q \int_A^B \vec{E} \cdot d\vec{r} \]
For a charge \( Q = 1 \, C \):
\[ W = -\int_A^B \vec{E} \cdot d\vec{r} \]

### Step 5: Substitute the Electric Field into the Integral
Substituting \( \vec{E} \) and \( d\vec{r} \) into the integral:
\[ \vec{E} \cdot d\vec{r} = (y\hat{i} + x\hat{j}) \cdot (dx\hat{i} + dy\hat{j}) = y \, dx + x \, dy \]

### Step 6: Set Up the Integral
Now we need to evaluate:
\[ W = -\int_{A}^{B} (y \, dx + x \, dy) \]

### Step 7: Determine the Path of Integration
We can parameterize the path from \( A \) to \( B \). We can move from \( (2, 2) \) to \( (4, 1) \) in two segments: first moving in the \( x \)-direction and then in the \( y \)-direction.

1. From \( (2, 2) \) to \( (4, 2) \) (constant \( y = 2 \)):
- Here, \( y = 2 \), \( dy = 0 \), and \( dx \) varies from 2 to 4.
- The integral becomes:
\[ W_1 = -\int_{2}^{4} (2 \, dx) = -2 \times (4 - 2) = -4 \]

2. From \( (4, 2) \) to \( (4, 1) \) (constant \( x = 4 \)):
- Here, \( x = 4 \), \( dx = 0 \), and \( dy \) varies from 2 to 1.
- The integral becomes:
\[ W_2 = -\int_{2}^{1} (4 \, dy) = -4 \times (1 - 2) = 4 \]

### Step 8: Combine the Work Done
Now, we combine the work done in both segments:
\[ W = W_1 + W_2 = -4 + 4 = 0 \]

### Final Answer
The work done in moving the charge from point A to point B is:
\[ \boxed{0 \, J} \]