00:01
We have a charge q1 located at origin.
00:04
So let this be our origin.
00:06
And this is charge q1.
00:08
Then we have another charge q2 located at 18 centimeters along the positive axis.
00:15
Let this be point r.
00:17
And then this is charge q2, which is 80 centimeters on the right hand side of the origin.
00:24
Okay.
00:25
And on top of this charge q2, right, we have a point, sorry, just a second.
00:38
On top of this charge q2, we have a point p, which is 13 .5 centimeters from point r.
00:47
Okay.
00:48
Now we have to find the electric field e at this point p due to q1.
00:56
So it will be like this.
01:00
Let this be even.
01:01
So the magnitude of charge q1 is equal to 0 .85 nanocoulombs 0 .85 sorry, 0 .85 nanocoulombs that is 0 .85 into into 10 raised to the power minus 9 coulombs.
01:29
Okay.
01:29
Now, this is it.
01:32
To find the electric field, we need to find this distance if origin is at o, then we have to find distance op.
01:39
So here we have a right angle triangle, right? so from pythagoras theorem, from pythagoras theorem, from pythagoras theorem, we have op square is equal to or square, sorry, or square plus pr square, right, which implies that op is equal to under root of the sum of squares of or and pr or rp square, same value, right.
02:32
So in this, if we substitute the value of or that is 18 centimeters and rp or pr, which is 13 .5 centimeters, we will have op is equal to 13 .5 sorry, 13 .5 square plus 18 square raised to the power 1 by 2 because this is under root, right.
03:04
So if we solve this calculation, we will find that op has a value that is the distance from the point p from the center is 22 .5 centimeters...