A uniform electric field of $5 \times 10^3\text{ N C}^{-1}$ is maintained in the positive Y-direction. Now a point charge of $2 \times 10^{-4}\text{ C}$ at rest is released from the origin. The kinetic energy attained by the charge when it is at a distance of $5\text{ m}$ from the origin is:
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Work is simply force times displacement. Group the coefficients together: $(2 \times 5 \times 5) = 50$, then combine the powers of ten: $10^{-4} \times 10^3 = 10^{-1}$. This gives $50 \times 0.1 = 5\text{ J}$ in a single mental calculation step.