Complete Question
Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 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.
Required:
What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?
Answer:
The velocity is
Explanation:
From the question we are told that
The magnitude of charge one is 
The magnitude of charge two 
The distance of separation is 
Generally the electric potential of the electron at the midway point is mathematically represented as

substituting values


Now when the electron is 10 cm = 0.10 m from charge 1 , it is (0.58 - 0.10 = 0.48 m ) m from charge two
Now the electric potential at that point is mathematically represented as

substituting values


Now the law of energy conservation ,
The kinetic energy of the electron = potential energy of the electron
i.e ![\frac{1}{2} * m * v^2 = [V_1 - V]* q](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%20m%20%2A%20v%5E2%20%20%3D%20%20%5BV_1%20-%20%20V%5D%2A%20q)
where q is the magnitude of the charge on the electron with value

While m is the mass of the electron with value 
![\frac{1}{2} * 9.11 *10^{-19} * v^2 = [ (3.67 - 1.8103) *10^{-8}]* 1.60 *10^{-19}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%2A%209.11%20%2A10%5E%7B-19%7D%20%2A%20v%5E2%20%20%3D%20%20%5B%20%283.67%20-%20%201.8103%29%20%2A10%5E%7B-8%7D%5D%2A%201.60%20%2A10%5E%7B-19%7D)
