00:01
So the diagram of the problem is shown as you can see that q1 is greater than q2 in magnitude and as this is plus and this is minus, so the electric field will be 0 at a point which is far from q1 and nearer to q2 because q2 is less.
00:26
Because for q1 the electric field will be in this direction and for q2 it will be in this direction so we will have actually two points.
00:34
One will lie within this distance 1 .5 meter and one will lie beyond q2, that is beyond 1 .5 meter.
00:45
Okay, so this is the thing.
00:46
So let's say x is the distance from q2, from q2 where the electric field is below, where the electric field is below.
01:10
So in that case we can write.
01:13
So let's say this is the point.
01:14
And this distance from q2 is x sorry this is x so from the diagram you can write 2 multiplied by 10 to the power minus 5 divided by x plus 1 .5 bode square is equal to 4 .2 multiplied by 10 to the power minus 6 divided by x square so if we saw now we will get x equal to 1 .27 meter and x equal to actually it is this x we have taken not percied like this so this is x.
02:20
So from here we are getting x equal to 1 .27 meter and x equal to 0 .47 meter.
02:36
So these are the distances from q2, therefore on the x axis, sorry, minus 0 .47 meters.
02:45
So the distances on the x axis where electric fields are 0 are 1 .5 plus 1 .27, that means equal to 1 .77 meter, and 1 .5 minus 0 .47 .7.
03:10
That is equal to 1 .03 meters.
03:16
So these are the two places...