1. +3.00μC point charge is at x=1.00cm and y=.750cm. -2.50nC point charge is at x=0 and y=-5.00mm. What will be the acceleration of an electron (Me=9.109X10^-31 kg) when it is at the origin (only force acting on the electron is the Coulomb force)? 2. K=8.988X10^9 N(m^2/c^2) Fc= K(qs qd/r^2sd)(direction rsd) I believe this is the electrostatic force formula? s=source d=detector. 3. This is my attempt at the problem. Fc=(K)(3.00X10^-6C)(2.50X10^-9C)/(2.5625X10^-4 m^2) [.624695i+.780869j] The 2.50X10^-9C was a negative to begin with but I redistributed the negative to the i and j hat direction to make them positive. |Fcsd|=.26306N θ=Tan^-1(.780869/.624695)=51.3degrees I cannot figure out how to get the acceleration of this electron? Do I use F=ma .26306N/(9.109X10^-31 kg) = 2.89X10^29 m/s^2 I am probably using the wrong formula to begin with can I please get some help with this.