00:01
Here we have to find out the value of the electric field intensity at a point.
00:04
Let's say point a is given that is equals to 1, 5 and 2 in a free space.
00:08
If the point charge is given that is equals to 6 pico column which is located at the point let's say point b which is 0, 0 and 1 and the uniform charge density is given to us that is equals to i which is equals to 18 newton column per meter along the x axis.
00:27
So all these values are given here.
00:29
So we need to find out the value of electric field intensity at the point a in the free space.
00:34
So from here we can say that let's say e1 is equals to q which is divided by the 4 pi epsilon not r square multiplied by the ar bar where ar is a unit vector.
00:45
So r bar from here is equals to point 1, 5 and 2 minus the point 0, 0 and 1.
00:52
So solving the term from here this become equals to 1 multiplied by the ax bar plus 5 which is multiplied by the a bar y bar plus az bar.
01:02
So the value of the vector a become equals to under root of 1 raised to the power 2 plus 5 raised to the power 2 plus 1 raised to the power 2.
01:10
Solving the term from here that from here is equals to 1 plus 25 plus 1.
01:14
So the value of r vector from here is to under root 27.
01:21
Now we are considering about e1 bar.
01:23
So e1 bar from here is equals to q which is divided by 4 pi epsilon not multiplied by the r square multiplied by the magnitude of ar vector.
01:31
So this from here is equals to q divided by q where q is 6 multiplied by the 10 raised to the power minus 6 which is divided by 4 pi multiplied by the epsilon not that is 8 .84 multiplied by the 10 raised to the power minus 12 that is further multiplied by the under root of 27 to its whole square ax bar plus 5 of ay bar plus az bar.
01:51
So this is the vector here...