Question

Two stationary positive point charges, charge 1 of magnitude 3.45nC and charge 2 of magnitude 1.80nC , are separated by a distance of 40.0cm . 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

I have solved this question earlier with different figures. Please workout using yours figures. If you need any further help just PM me. If I have helped you please rate me 5 stars first (before you rate anyone else).

two stationary positive point charges, charge 1 of magnitude 3.45nC and charge 2 of magnitude 1.80nC , are separated by a distance of 60.0cm . 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?

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)[(4*10^-9)+(1.95*10^-9)] / 0.285 V1 =187.895 V Initial kinetic energy of electron =KEi= zero Initial potential energy of electron =PEi= qV1 PEi =1.6*10^-19*187.895 PEi =300.63*10^-19 J Final potential =V2 =kQ1/r1 + k Q2/r2 r1 = 10 cm = 0.1 m r2 =57.0 - 10.0 = 47.0 cm = 0.47 m V2 =(9*10^9)[4*10^-9 /0.1 +1.95*10^-9/0.47] V2 =397.34 V Final kinetic energy of electron= KEf =(1/2)mv^2 Final potential energy of electron=PEf =qV2 PEf =635.75*10^-19 J KEf+PEf=KEi+PEi (1/2)mv^2 -635.75*10^-19 =zero - 300.63*10^-19 (1/2)mv^2 = 635.75*10^-19 - 300.63*10^-19 (1/2)mv^2 = [ 635.75 - 300.63 ]*10^-19 (1/2)mv^2 = 335.12 *10^-19 v = sq rt [2*(335.12*10^-19)/9,1*10^-31] v = sq rt [670.24*10^12/9.1] v = sq rt 73.653*10^12 v =8.58*10^6 m/s The speed v(final) of the electron when it is 10.0 cm from charge 1 is 8.58*10^6 m/s