If an electron is placed in an electric field with a potential difference of (1.35x10^2) V, the maximum speed the electron can attain is?

2 Answers
Jan 7, 2018

rArr v_f = sqrt((2eDeltaV)/m)

Explanation:

Use the work-energy principle or conservation of energy.

Let
V = electric potential
U = potential energy = qV
K = kinetic energy = 1/2mv^2
v_i= initial velocity = 0
v_f = final velocity
e = electron charge = 1.602 xx10^-19 C
q = charge = -e
m= mass of the electron = 9.11 xx10^-31

Most people are familiar with conservation of energy:

Delta(U +K) =0 rArr DeltaK= - DeltaU

rArr 1/2mv_f^2- 1/2mv_i^2 = -qDeltaV
rArr 1/2mv_f^2 - 0 = -(-e)DeltaV
rArr v_f = sqrt((2eDeltaV)/m)

By estimation,
v_f ~~10^7 m/s

Jan 7, 2018

6.89*10^6ms^(-1)

Explanation:

Here the electron will gain kinetic energy while losing electric potential energy. From this we can get:
Delta(1/2mv^2)=Delta(QV)

We can rearrange that equation:
mv^2=2eV
v^2=(2eV)/m
v=sqrt((2eV)/m

We know that:
e=1.6*10^(-19)color(white)(l)C
V=1.35*10^2color(white)(l)V
m=9.11*10^(-31)kg
v=sqrt((2(1.6*10^(-19))(1.35*10^2))/(9.11*10^(-31)))~~6886248.41~~6.89*10^6ms^(-1)