Two stationary positive point charges, charge 1 of magnitude 3.35 nC and charge 2 of magnitude 1.75 nC , are separated by a distance of 41.0 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 vfinal of the electron when it is 10.0 cm from charge 1?

Pe = 1/2k(x)^2  and .05n/m = .05x10^-9m = x 

Distance covered, s = 8ft - 4ft = 4 ftinitial velocity, u = 34 ft/sthe acrobat is moving upwards and therefore, a = -g = -9.81 m/s^2 = 32.19 ft/s^2applying equation of motion; s = ut -1/2gt^2then,4 = 34t -1/2*32.19t^24 = 34t - 16.095t^2rearranging; -16.095t^2 + 34t - 4 = 0solving the quadratic equation; t = {-34 +/- sqrt [34^2 - 4(-16.)]}/2(-16.095) = 1.06 +/- 0.93 = 0.13 or 1.99the acrobat will take 0.13 seconds to be at 8 ft above the ground. additionally, once at the maximum height, the acrobat will move down and be at the same spot (that is height) after 1.99 seconds moving down.

Carbon dioxide because it consists of one type of molecule.

Can you give more detail