A proton is traveling horizontally to the right at 4.20×106 m/s. Find (a)the magnitude and (b) direction (counterclockwise from the left direction) of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.40 cm. Part C: How much time does it take the proton to stop after entering the field? What minimum field ((Part D)magnitude and (Part E)direction) would be needed to stop an electron under the conditions of part (a)? ATTEMPT: PART A: i used two formulas. vf^2=V0^+2a(deltaX) and E=F/q. i solved the first one for a and plugged it into the second one getting -2.7E6 N/C. i put a negative sign because the field must attract the proton. but i'm not sure if this is right. PART B: im not sure what the question means by "counterclockwise from the left direction" but i think the direction is due left. which would normally be 180 degrees. PART C: using the equation vf=v0+at i plugged in my values found above and got t= 1.62E-8 seconds. PART D: i used the same method in D except i changed the mass to the mass of an electron. i got 14878.75 N/C PART E: same for PART B are these right?