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CGP EDU Academic Team
Published on: September 12, 2026
A mass $m = 20 \, g$ has a charge $q = 3.0\,mC$ . It moves with a velocity of $20 \, m/s$ and enters a region of electric field of $80 \ \mathrm{N/C}$ in the same direction as the velocity of the mass. The velocity of the mass after 3 seconds in this region is
Text Solution
Verified by ExpertsThe correct answer is:
B
$a = \frac{QE}{m} = \frac{3 \times 10^{-3} \times 8 0}{20 \times 10^{-3}} = 12 m / sec^{2}$
Hence v = u + at ⇒ ⇒ v = 20 + 12 × × 3 = 56 m/s.
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