An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3 us (microsecond??) after it is released. Part A What is the magnitude of the electric field? How would I calculate the magnitude of the electric field? I am not sure which forumla to use, but I am guessing E = F/q E= F/q also equals 1/(4pi(8.854x10^-12)) * Q/r^2 If that is the right equation to use, then I know we have r already (4.50m) but how do we solve for Q? Part B What is the direction of the electric field? Upward or Downward? Since it says the electron moves upward after released, will that mean the direction of the electric field is upward also? Thank you for your help.