Two stationary positive point charges, charge 1 of magnitude 3.65 nC and charge

Question

Two stationary positive point charges, charge 1 of magnitude 3.65 nC and charge 2 of magnitude 2.00 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 vfinal of the electron when it is 10.0 cm from charge 1?

Express your answer in meters per second.

Two stationary positive point charges, charge 1 of magnitude 3.65 nC and charge 2 of magnitude 2.00 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 vfinal of the electron when it is 10.0 cm from charge 1?

Express your answer in meters per second.

Explanation / Answer

given

distance b/w 2 fixed charges = d = 40 cm

q1 = 3.65 nC

q2 = 2 nC

by the mechanical energy conservation we can write

KEi + PEi = KEf + PEf

we know that PEi = ke q1/ (d/2) + k e q2/(d/2)

= 2ke (q1+q2)/d

= 2* 9*109* (-1.6 * 10-19)(3.65+2) * 10-9 / 0.4   

= -4.068 * 10-17 J

PEf = keq1 / 0.1 + keq2/ 0.3 = 9*109 * (-1.6 * 10-19) *3.65* 10-9 / 0.1 +   9*109 * (-1.6 * 10-19 )*2* 10-9 /0.3

= -5.256 * 10-17 - 0.96 * 10-17 J

=  -6.216 * 10-17 J

let velocity at that time = v m/s

so KEf = 1/2 m v2

so from   

KEi + PEi = KEf + PEf

0 - 4.068 * 10-17 = 1/2 m v2 - 6.216 * 10-17

2.148 *  10-17 = 1/2 * 9.1*10-31 v2

v = 6.87 * 106 m/s

answer

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