Question

Two stationary positive point charges, charge 1 of magnitude 3.40 nC and charge 2 of magnitude 1.75 nC , are separated by a distance of 40.0 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 *v*final of the electron when
it is 10.0 cm from charge 1?

Answer #1

q2 = + 1.75*10^-9 C

Separation by distance = s = 40.0cm =0.4 m.

Electric potential energy of electron at 0.20 m =potential energy
due to q1 + potential energy due to q2

Electric potential energy of electron at 0.20 m
=-9*10^9(1.6*10^-19)[ 3.40*10^-9 + 1.75*10^-9 ] /0.20

Electric potential energy of electron at 0.20 m =-9*(1.6*10^-19) [
3.40+ 1.75 ] /0.20

Electric potential energy of electron at 0.25 m = -
3.708*10^-17J

Electric potential energy of electron at 0.10 m
=-9*10^9(1.6*10^-19)*10^-9[ 3.40/0.1 + 2.00/0.3]

Electric potential energy of electron at 0.10 m =-5.856 *10^ -17
J

Change in PE =2.148*10^-17 J

KE =2.148*10^-17 J

Speed = v = sq rt [2KE/m ]

Speed = v = sq rt [2*2.148*10^-17 /9.1*10^ -31 ]

Speed = v = 6.870*10^6 m/s

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ASAP!!
I'll rate it as soon as the answer works!
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