Question

Two stationary positive point charges, charge 1 of magnitude 3.20nC and charge 2 of magnitude 2.00nC , are separated by a distance of 42.0cm . 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 v-final of the electron when it is 10.0cm from charge 1?

Answer 1

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

Q2=1.95 nC= 1*9510^-9 C

Total seperation =D=57 cm=0.57 m

Distance of either charge from the mid point=r1= D / 2 = 0.57 /2= 0.285 m

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

V1=k[Q1+Q2] / r

V1 =(9*10^9)[(3.2*10^-9)+(2*10^-9)] / 0.21

V1 =222.857 V

Initial kinetic energy of electron =KEi= zero

Initial potential energy of electron =PEi= qV1

PEi =1.6*10^-19*222.857

PEi =356.571*10^-19 J

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

r1 = 10 cm = 0.1 m

r2 =42.0 - 10.0 = 32.0 cm = 0.32 m

V2 =(9*10^9)[3.2*10^-9 /0.1 +2*10^-9/0.32]

V2 =344.25 V

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

Final potential energy of electron=PEf =qV2

PEf =550.08*10^-19 J

KEf+PEf=KEi+PEi

(1/2)mv^2 -550.08*10^-19 =zero - 356.571*10^-19

(1/2)mv^2 = 550.08*10^-19 - 356.571*10^-19

(1/2)mv^2 = [ 550.08 - 356.571 ]*10^-19

(1/2)mv^2 = 194.229.12 *10^-19

v = sq rt [2*(194.229.12 *10^-19 )/9.1*10^-31]


v =6.533*10^6 m/s

The speed v(final) of the electron when it is 10.0 cm from charge 1 is v =6.533*10^6 m/s m/s