A charge of $9 μC$ is given a displacement of $\vec r= (4\hat i +5\hat j) m$ in an electric field $\vec E= (5\hat i+4\hat j) N/C$. The work done in the process is

$3.6 × 10^{-4} J$

The correct answer is Option (4) → $3.6 × 10^{-4} J$

Given:

Charge $q = 9 \, \mu C = 9 \times 10^{-6} \, C$

Displacement $\vec{r} = 4 \hat{i} + 5 \hat{j} \, m$

Electric field $\vec{E} = 5 \hat{i} + 4 \hat{j} \, N/C$

Work done by electric field:

$W = q \, \vec{E} \cdot \vec{r}$

Dot product:

$\vec{E} \cdot \vec{r} = (5)(4) + (4)(5) = 20 + 20 = 40$

Therefore:

$W = 9 \times 10^{-6} \times 40 = 3.6 \times 10^{-4} \, J$

Answer: $W = 3.6 \times 10^{-4} \, J$