Understanding the Concept:
When a charged body moves through a uniform electric field $\vec{E}$, it experiences a constant force $\vec{F} = q\vec{E}$. The work done by this electric field over a displacement vector $\vec{d}$ parallel to the lines of force equals the system's kinetic energy gain:
\[
K = W = F \cdot d = q \cdot E \cdot d
\]
Step 1: Verify alignment and compute work value components.
We are given:
Electric Field magnitude, $E = 5 \times 10^3\text{ N C}^{-1}$ along $+\hat{j}$
Charge value, $q = 2 \times 10^{-4}\text{ C}$
Distance along movement path, $d = 5\text{ m}$
Since the charge is positive and released from rest, it moves directly along the field lines parallel to the Y-axis. The work done is:
\[
W = (2 \times 10^{-4}\text{ C}) \times (5 \times 10^3\text{ N C}^{-1}) \times (5\text{ m})
\]
Step 2: Evaluate numerical multiplication terms.
\[
W = 2 \times 5 \times 5 \times 10^{-4} \times 10^3 = 50 \times 10^{-1} = 5\text{ J}
\]
Since the particle started from rest ($K_{\text{initial}} = 0$), the kinetic energy attained is equal to the work done, which is $5\text{ J}$.