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
Answer & explanation
Correct answer: option 4
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: $W = 3.6 \times 10^{-4} \, J$