Two particles with positive charges q1 and q2 are separated by a distance s. Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges zero?

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The chemical reactions lab we did is typically called a "copper plating lab,  because the acid solution cleaned the pennies and made the copper look shinier;   the acid solution cleaned the paperclips and made them look like pennies;   &   the copper from the pennies was removed, and then added to the paperclips. 

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.