Question
Physics Question on electrostatic potential and capacitance
An electron initially at rest, is accelerated through a potential difference of 200 volt, so that it acquires a velocity 8.4×106ms−1, the value of e/m of electron will be
A
1.76×1011Ckg−1
B
3.76×1012Ckg−1
C
4×1011Ckg−1
D
0.76×1011Ckg−1
Answer
1.76×1011Ckg−1
Explanation
Solution
Maximum kine KEmax=eV
or 21mv2=eV
=212×102(8.4×106)2=1.764×1011Ckg−1