Aconstant electric field with magnitude 1.50 ✕ 103 n/c is pointing in the positive x-direction. an electron is fired from x = −0.0200 m in the same direction as the electric field. the electron's speed has fallen by half when it reaches x = 0.170 m, a change in potential energy of 4.56 ✕ 10−17 j. the electron continues to x = −0.210 m within the constant electric field. if there's a change in potential energy of −9.12 ✕ 10−17 j as it goes from x = 0.170 m to x = −0.210 m, find the electron's speed (in m/s) at x = −0.210 m. hint notice the electron turned around at some point. when the electron's speed is half its initial value, you are given the change in electric potential energy. the sum of δu and δk at this point is a conserved quantity, and is equal to zero, so δu + δk will be the same at all points along the electron's trajectory. you are given δu at the point of interest, so write δk in terms of velocities, substitute the velocity at the second point with the initial velocity divided by 2, simplify your equation, and solve for vf.