Charge `Q` is given a displacement `vec r = a hat i + b hat j` in an electric field `vec E = E_1 hat i + E_2 hat j`. The work done is.

Options

A. `Q(E_1 a + E_2 b)`
B. `Q sqrt((E_1 a)^2 + (E_2 b)^2)`
C. `Q (E_1 + E_2) sqrt(a^2 + b^2)`
D. `Q sqrt((E_1^2 + E_2^2)^2) sqrt (a^2 + b^2)`

Explanation

To solve the problem of calculating the work done when a charge \( Q \) is displaced in an electric field \( \vec{E} \), we can follow these steps:

### Step 1: Understand the formula for work done
The work done \( W \) by a force \( \vec{F} \) when moving an object through a displacement \( \vec{r} \) is given by:
\[ W = \vec{F} \cdot \vec{r} \]

### Step 2: Identify the force acting on the charge
The force \( \vec{F} \) acting on a charge \( Q \) in an electric field \( \vec{E} \) is given by:
\[ \vec{F} = Q \vec{E} \]

### Step 3: Substitute the electric field
Given that the electric field \( \vec{E} \) is:
\[ \vec{E} = E_1 \hat{i} + E_2 \hat{j} \]
We can express the force as:
\[ \vec{F} = Q (E_1 \hat{i} + E_2 \hat{j}) = Q E_1 \hat{i} + Q E_2 \hat{j} \]

### Step 4: Identify the displacement vector
The displacement \( \vec{r} \) is given as:
\[ \vec{r} = a \hat{i} + b \hat{j} \]

### Step 5: Calculate the dot product
Now, we need to compute the dot product \( \vec{F} \cdot \vec{r} \):
\[ \vec{F} \cdot \vec{r} = (Q E_1 \hat{i} + Q E_2 \hat{j}) \cdot (a \hat{i} + b \hat{j}) \]

Using the properties of the dot product:
- \( \hat{i} \cdot \hat{i} = 1 \)
- \( \hat{j} \cdot \hat{j} = 1 \)
- \( \hat{i} \cdot \hat{j} = 0 \)
- \( \hat{j} \cdot \hat{i} = 0 \)

We can simplify the dot product:
\[ \vec{F} \cdot \vec{r} = (Q E_1 a) + (Q E_2 b) = Q E_1 a + Q E_2 b \]

### Step 6: Write the final expression for work done
Thus, the work done \( W \) is:
\[ W = Q E_1 a + Q E_2 b \]

### Conclusion
The work done when the charge \( Q \) is displaced in the electric field \( \vec{E} \) is given by:
\[ W = Q (E_1 a + E_2 b) \]