Electrostatics Question 439

Question: The acceleration of an electron in an electric field of magnitude 50 V/cm, if e/m value of the electron is 1.76×1011 C/kg, is [CPMT 2001]

Options:

A) 8.8×1014 m/sec2

B) 6.2×1013 m/sec2

C) 5.4×1012 m/sec2

D) Zero

Show Answer

Answer:

Correct Answer: A

Solution:

a=eEma=1.76×1011×50×102

=8.8×1014m/sec2



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