ΔV is constant for both situations. So if I use ΔU = qΔV, when charge increases, ΔU has to increase also, which means kinetic energy has decreased. Since KE has decreased, and we know that KE = 1/2 mv^2, the speed of the second particle will be less than the speed of the second particle. I hope...
Homework Statement
The four points shown in the attached picture are near a positively charged rod (shaded circle). Points W and Y are equidistant from the rod, as are points X and Z. A charged particle with mass 3.0 x 10^-8 kg is released from rest at point W and later observed to pass...