Two stationary positive point charges, charge 1 of magnitude 3.20 nC and charge

Question

Two stationary positive point charges, charge 1 of magnitude 3.20 nC and charge 2 of magnitude 1.65 nC , are separated by a distance of 49.0 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.

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Two stationary positive point charges, charge 1 of magnitude 3.20 nC and charge 2 of magnitude 1.65 nC , are separated by a distance of 49.0 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.

View Available Hint(s)

nothing

SubmitPrevious Answers

Incorrect; Try Again; 5 attempts remaining

Explanation / Answer

Q1=4.00 nC= 3.2*10^-9 C

Q2=1.65 nC= 1.65*10^-9 C

Total seperation =D=49 cm=0.49 m

Distance of either charge from the mid point=r1= D / 2 = 0.49 /2= 0.245 m

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

V1=k[Q1+Q2] / r

V1 =(9*10^9)[(3.2*10^-9)+(1.65*10^-9)] / 0.245

V1 =178.163 V

Initial kinetic energy of electron =KEi= zero

Initial potential energy of electron =PEi= qV1

PEi =1.6*10^-19*178.163

PEi = 285.06*10^-19 J

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

r1 = 10 cm = 0.1 m

r2 =49.0 - 10.0 = 39.0 cm = 0.39 m

V2 =(9*10^9)[3.2*10^-9 /0.1 +1.65*10^-9/0.39]

V2 =326.08 V

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

Final potential energy of electron=PEf =qV2

PEf =521.72*10^-19 J

KEf+PEf=KEi+PEi

(1/2)mv^2 -521.72*10^-19 =zero - 326.08*10^-19

(1/2)mv^2 = 195.64 *10^-19

v = sq rt [2*(195.64*10^-19)/9.1*10^-31]

v = 6.55*10^6 m/s

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