00:01
In this problem we have been given that there are two point charges which are placed on the x -axis.
00:07
One of the charge is present at 3 cm.
00:11
So let's say this is the origin here.
00:15
So at 3 centimeter to the right, one of the particle is present and that has the charge of 5 .07 coulum.
00:27
And there is another particle which is situated at plus 9 centimeter.
00:32
So let's just extend this x -axis.
00:38
So here the particle is positioned at 9 centimeter, and the charge here is minus 23 column.
00:47
So we need to figure out the total electric field in terms of magnitude as well as the direction at first x equals to 0 centimeter.
00:58
So here we need to figure out the electric field.
01:01
So we are going to apply here the expression to get the electric field with respect to point charge using kq by r square.
01:09
So at the origin, the electric field, because of the charge placed at 3 centimeter, that would be in the left direction.
01:17
And that would be, let's say this is e1, so we can get the value of e1 as k times the charge 5 .01 over r square.
01:29
That's 0 .03 meters square.
01:32
So when we simplify here, we multiply 9 with 5 .07 and divide.
01:38
With 0 .03 that comes out to be 5 .1 into 10 raised to 13 newtons per column.
01:49
And the electric field due to the charge placed at 9 cm, that would be in the right direction.
01:55
So that will be 9 times 10 raise to 9 into the magnitude of charge divided by 0 .09 whole square.
02:04
So this we multiply the numerator and divide by square of point 0 .0.
02:10
So that comes out to be 2 .6 into 10 raised to 13 newtons per column.
02:19
So this comes out to 2 .6 into 10 raise to 13 newtons per column...