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)#