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\times 10^{11} $ C/kg, is [CPMT 2001]

Options:

A) $ 8.8\times 10^{14} $ m/sec2

B) $ 6.2\times 10^{13} $ m/sec2

C) $ 5.4\times 10^{12} $ m/sec2

D) Zero

Show Answer

Answer:

Correct Answer: A

Solution:

$ a=\frac{eE}{m}\Rightarrow a=1.76\times 10^{11}\times 50\times 10^{2} $

$ =8.8\times 10^{14}m/sec^{2} $



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