An electron is released from rest in a uniform electric field and accelerates to the north at a rate of $105 \textrm{ m/s}^2$. Find the magnitude and direction of the electric field.

VIDEO TRANSCRIPT

This is Giancoli Answers with Mr. Dychko. So this electron experience a force directed north in which case the electric field must be directed south because electric field points in the direction of force applied on a positive charge. And so being a negative charge, the electric field a point in the opposite direction to the force experienced by an electron and presumably the electrostatic force is the only one being applied in this electron in which case it's the net force. The static force is a net force and that's going to be mass times acceleration. And that equals electrostatic force and we can solve this for E by dividing both sides by q. So the electric field strength is ma over q. So that's 9.11 times ten to the minus 31 kilograms, mass of an electron, times a 105 meters per second squared acceleration divided by the charge on an electron, 1.6 times ten to the minus 19 coulombs which gives 5.79 times ten to the minus ten Newtons per coulombs directed south for the electric field.