Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 30.0 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. What is the speed vfinal of the electron when it is 10.0 cm from charge 1?

-6.95*10^-17 J

Explanation:

Given that

Magnitude of first charge, q1 = 3.9*10^-9 C

Magnitude of second charge, q2 = 1.85*10^-9 C

Distance between the two charges, r = 30 cm = 0.3 m

The potential energy of point charges is given as

U = kqq•/r, where

k = constant of proportionality, 8.99*10^9 Nm²/C²

See attachment for solution.

At the part I tagged B, the distance of the second charge 30 cm - 10 cm, and that is how I got my 20 cm or 0.2 m as I used it.

The final answer would have been

U = U1 + U2

U = -5.62*10^-17 + -1.33*10^-17

U = -6.95*10^-17 J

amplitude

hope this