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A proton (1.67e-27 kg) sits in a field of 2.0e4 N/C. Its acceleration is about?
A1.9 x 10^12 m/s^2
B1.9 x 10^9 m/s^2
C3.5 x 10^15 m/s^2
D1.2 x 10^-8 m/s^2
Answer & Solution
Correct answer: A. 1.9 x 10^12 m/s^2
a = eE/m = 1.6e-19 x 2.0e4 / 1.67e-27, which gives roughly 1.9 x 10^12 m/s^2.
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