00:02
Problem number 24, we're given the magnitude of the electric force acting on a charge as 8 .4 newton's.
00:10
It is also the direction of this force is given to us as downward.
00:16
And also the charge itself is negative 8 .8 microcolones.
00:23
So we're asked the magnitude and direction of the e -fuel.
00:32
Acting on this chart.
00:36
So as you know, the electric field equals two of the force divided by the charge.
00:45
In our case, let's write this f sub e vector as the magnitude f of e times the direction of this force.
00:57
So let's pick downward conventionally to be, downward to be in the negative wide and upward to be in the positive wide direction.
01:07
So then the unit vector that represents the negative wide direction is negative j hat divided by q again, right? so f of e, the magnitude is 8 .4 newtons...