Question

Two stationary positive point charges, charge 1 of magnitude 3.75 nC and charge 2 of magnitude 1.90 nC , are separated by a distance of 35.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. Part A What is the speed vfinal of the electron when it is 10.0 cm from charge 1? Express your answer in meters per second.

Answer 1

Given that -

Q1= 3.75 nC = 3.75 *10^-9 C

Q2 =1.90 nC= 1.90 *10^-9 C

Distance between them =D=35 cm=0.35 m

Distance of either charge from the mid point=r1= D / 2 = 0.35 /2= 0.175 m

Now, the initial potential (at mid point) = V1 =kQ1/r +kQ2/r

V1=k[Q1+Q2] / r

V1 =(9*10^9)[(3.75*10^-9)+(1.90*10^-9)] / 0.175

V1 =290.6 V

Initial kinetic energy of electron =Kei = zero

Initial potential energy of electron =Pei = qV1

PEi =1.6*10^-19*290.6

PEi =465.0*10^-19 J

Final potential =V2 =kQ1/r1 + k Q2/r2

r1 = 10 cm = 0.1 m

r2 =35.0 - 10.0 = 25.0 cm = 0.25 m

V2 =(9*10^9)[3.75*10^-9 /0.1 +1.90*10^-9/0.25]

V2 = 9*[37.5 + 7.60] = 405.9 V

Final kinetic energy of electron= KEf =(1/2)mv^2

Final potential energy of electron=PEf =qV2

PEf =1.6 *10^-19 * 405.9 = 649.4*10^-19 J

Now -

KEf+PEf=KEi+PEi

(1/2)mv^2 -649.4*10^-19 =zero - 465.0*10^-19

(1/2)mv^2 = 649.4*10^-19 – 465.0*10^-19

(1/2)mv^2 = [ 649.4 – 465.0 ]*10^-19

(1/2)mv^2 = 184.4 *10^-19

v = sq rt [2*(184.4*10^-19)/9.1*10^-31]

v = sq rt [184.4*10^12/9.1]

v = sq rt [20.26*10^12] = 4.50 x 11^6 m/s

So, the final speed of the electron = 4.50 x 10^6 m/s.