Question

Two stationary positive point charges, charge 1 of magnitude 4.00nC and charge 2 of magnitude 1.70nC , are separated by a distance of 39.0cm . 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 10.0cm from charge 1?

Answer 1

Q1=4.00 nC= 4*10^-9 C

Q2=1.70 nC= 1*7010^-9 C

Total seperation =D=39 cm=0.39 m

Distance of either charge from the mid point=r1= D / 2 = 0.39 /2= 0.195 m

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

V1=k[Q1+Q2] / r

V1 =(9*10^9)[(4*10^-9)+(1.70*10^-9)] / 0.195

V1 =263.076 V

Initial kinetic energy of electron =KEi= zero

Initial potential energy of electron =PEi= qV1

PEi =1.6*10^-19*263.076

PEi =4.209*10^-17 J

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

r1 = 10 cm = 0.1 m

r2 =39.0 - 10.0 = 29.0 cm = 0.29 m

V2 =(9*10^9)[4*10^-9 /0.1 +1.70*10^-9/0.47]

V2 =392.55 V

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

Final potential energy of electron=PEf =qV2

PEf =1.6*10^-19*392.55 J = 6.28 x 10^-17 J

KEf+PEf=KEi+PEi

(1/2)mv^2 -6.28 x 10^-17 =zero - 4.209*10^-17

(1/2)mv^2 = 6.28 x 10^-17 - 4.209*10^-17

(1/2)mv^2 = 2.071 *10^-17  

v = sqrt [2*(2.071 *10^-17)/9.1*10^-31]



v =6746590.5 m/s (answer)