Question

An electron is released from rest in a uniform electric field of magnitude 2.16 x 10^4 N/C. Calculate the acceleration of the electron. (Ignore gravitation.) Number ______ Units _________

Answer 1

Answer:

Acceleration of the electron will be $0.379\times 10^{16}m/sec^2$          

Explanation:

We have given electric field $E=2.16\times 10^4N/C$

Charge on electron $q=1.6\times 10^4C$

So force in the electron $F=qE=1.6\times 10^{-19}\times 2.16\times 10^{4}=3.456\times 10^{-15}N$

Mass of electron $m=9.11\times 10^{-31}kg$

Now according to newton second law

F = ma, here m is mass and a is acceleration

So $3.456\times 10^{-15}=9.11\times 10^{-31}a$

$a=0.379\times 10^{16}m/sec^2$

So acceleration of the electron will be $0.379\times 10^{16}m/sec^2$