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} $
