Charge Q is given a displacement `r=ahati+bhatj` in electric field `E=E_1hati+E_2hatj`. The work done is

Options

A. `Q(E_1a+E_2b)`
B. `Qsqrt((E_1a)^2+(E_2b)^2)`
C. `Q(E_1+E_2)sqrt(a^2+b^2)`
D. `Qsqrt(E_1^2+E_2^2) sqrt(a^2+b^2)`

Explanation

To find the work done when a charge \( Q \) is displaced in an electric field \( \mathbf{E} \), we can follow these steps:

### Step 1: Understand the Work Done in an Electric Field
The work done \( W \) when a charge \( Q \) is moved in an electric field \( \mathbf{E} \) over a displacement \( \mathbf{r} \) is given by the formula:
\[ W = Q \Delta V \]
where \( \Delta V \) is the change in electric potential.

### Step 2: Identify the Electric Field and Displacement Vectors
The electric field is given as:
\[ \mathbf{E} = E_1 \hat{i} + E_2 \hat{j} \]
The displacement vector is given as:
\[ \mathbf{r} = a \hat{i} + b \hat{j} \]

### Step 3: Calculate the Change in Potential \( \Delta V \)
The change in potential \( \Delta V \) can be found using the dot product of the electric field vector and the displacement vector:
\[ \Delta V = \mathbf{E} \cdot \mathbf{r} \]
Calculating this dot product:
\[ \Delta V = (E_1 \hat{i} + E_2 \hat{j}) \cdot (a \hat{i} + b \hat{j}) = E_1 a + E_2 b \]

### Step 4: Substitute \( \Delta V \) into the Work Done Formula
Now substitute \( \Delta V \) into the work done formula:
\[ W = Q \Delta V = Q (E_1 a + E_2 b) \]

### Step 5: Final Expression for Work Done
Thus, the work done \( W \) when the charge \( Q \) is displaced in the electric field is:
\[ W = Q (E_1 a + E_2 b) \]

### Conclusion
The work done in moving the charge \( Q \) in the electric field \( \mathbf{E} \) over the displacement \( \mathbf{r} \) is given by:
\[ W = Q (E_1 a + E_2 b) \]

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