Question

# Two stationary positive point charges, charge 1 of magnitude 3.70 nC and charge 2 of magnitude...

Two stationary positive point charges, charge 1 of magnitude 3.70 nC and charge 2 of magnitude 1.50 nC , are separated by a distance of 46.0 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.

What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

Express your answer in meters per second.

Q1=4.00 nC= 3.7*10^-9 C

Q2=1.95 nC= 1.5 *10^-9 C

Total seperation =D=46 cm=0.46 m

Distance of either charge from the mid point=r1= D / 2 = 0.46 /2= 0.23 m

Initial potential (at mid point) = V1 =kQ1/r +kQ2/r

V1=k[Q1+Q2] / r

V1 =(9*10^9)[(3.7*10^-9)+(1.5*10^-9)] / 0.23

V1 =203.48 V

Initial kinetic energy of electron =KEi= zero

Initial potential energy of electron =PEi= qV1

PEi =1.6*10^-19*203.48

PEi =325.56*10^-19 J

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

r1 = 10 cm = 0.1 m

r2 =46.0 - 10.0 = 36.0 cm = 0.36 m

V2 =(9*10^9)[3.7*10^-9 /0.1 +1.5*10^-9/0.36]

V2 =370.5 V

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

Final potential energy of electron=PEf =qV2

PEf =592.8*10^-19 J

KEf+PEf=KEi+PEi

(1/2)mv^2 – 592.8*10^-19 =zero – 325.56*10^-19

(1/2)mv^2 = 592.8*10^-19 – 325.56*10^-19

(1/2)mv^2 = 267.24*10^-19

v = sq rt [2*(267.24*10^-19)/9.1*10^-31]

v = 7.66*10^6 m/s