00:01
For this problem on the topic of electric fields, we are told that a rod of length l that is uniformly charged with the total charge of q lies along the y -axis.
00:09
It is lying from y is equal to 0 to y is equal to l, and we want to find an expression for the electric field at the point d -0.
00:19
Now, the charge q is uniformly distributed along the length of the rod l, and so the rod has a linear charge density lambda, which is q over l.
00:28
The electric field at a position x is equal to d can be calculated by integrating over the differential electric field due to the differential charge in the rod.
00:36
The electric field differential component, de, is equal to kdq over r squared, where the differential is along the y axis and r is equal to the square root of d squared plus y squared.
00:59
Now the x and y components of the field must be considered individually.
01:03
The x component of the field differential is given by minus de cosine theta and the y component by de sine theta.
01:15
And so we'll first find the x component of the electric field differential, dex, which is k.
01:25
Dq over r squared cosine cosine theta, which we can write as k lambda.
01:37
Dy over r squared cosine theta, which is k times q, dy divided by lr squared times the cosine of theta.
01:59
And we can write this as k times the total charge q times dy divided by l into an r squared can be replaced by d...