Traveling proton and electric field problem

[kirby2]

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: I am 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?

 

[TSny]

Looks good except for the direction of the E field for the case of the electron. (The electron has a negative charge.)

 

[Chandra Prayaga]

All uncertainties of negative signs are removed by drawing a picture. Draw the diagram with the x axis, show the proton moving to the right (in the + x direction). Now ask the question, what direction should the force be to stop the proton? That will also give the direction of the electric field, and then the sign will be obvious. This diagram should be drawn before starting on any formulas.

 

[LastScattered1090]

PART B: I am 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.

@kirby2 Counterclockwise from the left means start facing due left, and measure all vector directions relative to this reference direction (with positive angles being rotations CCW from this direction). So if the electric field needs to be due left, it would be at an angle of 0 degrees, according to this coordinate convention. It's not a usual coordinate convention, which would measure things CCW from +x (right) or from +y (up). But it does seem to be what the problem is calling for here.