00:01
So in this problem, we're asked first without this q3 charge, what is the potential and electric field at the origin? and then once we add this charge, what is the electric force on the third charge? and what is the electric potential energy of the third charge? so first, we can find the potential and the electric field at the origin.
00:32
These two point charges and we can use the equations that the potential is equal to kq over r the electric field is equal to kq over r squared where k is this constant q is the magnitude of the charge of the particle in question and e is r squared or r squared is the distance between the point and the charge and then we can use superposition so the charge or the rather the potential and electric field at the origin is going to be given by e of q1, e of q2, and b of q1 plus v of q2, respectively.
01:54
So the first thing the question asks is the electric field.
02:04
So for an electric field.
02:15
So since q1 is directly above our point and positive, right? so q1 is directly above our point and positive.
02:40
So its e field at the origin should point straight downwards in the negative x.
02:47
Meanwhile, q2 at that point, its e field should point directly to the left.
02:59
Which in this case is our negative x and this should point in the negative y direction so or actually sorry excuse me because this is a negative charge it wants to pull on things and so it's going to pull in the positive x direction that's my fault so i'll just finish the sentence quickly since q1 is directly above our point in positive it points downwards and we have the electric field of q1 is equal to k times q1 over the distance between it and the origin squared in the negative y hat direction.
04:13
And that's going to equal k is 8 .99 times 10 to the 9th.
04:24
Q1 is one nanoculum, so 1 times 10 to the negative 9 coulums.
04:34
And our distance given in the problem is 1 meter squared.
04:46
And so this is going to equal 8 .99 newtons per coulom.
05:00
And that should be negative in the y hat direction.
05:04
And the electric field of q2 is going to equal k times q2 over its distance squared.
05:17
And this time, right, remember this should be in the positive x hat direction.
05:27
And this is going to equal 8 .99.
05:46
And the reason why i paused here is because i realized that in choosing the direction, i actually need to make this the magnitude of q2 because we still want our electric field to point in the right direction.
06:04
And so that means that we need to make the magnitude, this is the magnitude of q, which is going back up two nanocrylombs, 2 times 10 to the negative 9 over 2 meters squared.
06:23
So 8 .99 times 2 by 2.
06:30
Square.
06:32
So this is going to give 4 .495 newton's per kulam, x hat direction.
06:50
And so we're going to have the total is going to equal 4 .495 in the x -hat direction minus 8 .999 in the y -hat direction, and this is all newton's per cullum.
07:24
The potential, right, doesn't have a direction.
07:28
So for potential, we have v equals kq over r...