Question

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 *v*final of the electron when it is
10.0 cm from charge 1?

Express your answer in meters per second.

Answer #1

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**

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