00:01
In this question, you want to find an electric field at point p due to the uniformly charged rod.
00:09
Okay, so this is a question on finding electric field due to continuous charge distribution.
00:17
So we are going to chop the rod into smaller pieces.
00:21
Okay, so suppose we have here the x or dq and then at position x away, from the origin and due to this dq is going to generate or produce a d .e at point p that's pointing in this direction.
00:47
Okay, and then based on the diagram you can see that since the rod is symmetrical about the y axis, so you can tell that the horizontal components that are going to cancel out.
01:01
Okay, so we are going to find a vertical components of the de.
01:07
Okay, so we need this angle data, and this is also angle data.
01:12
And we need this distance for the electric field calculations, and this distance will be x squared plus r square using pyrtegris theorem.
01:23
And then later we need the side data.
01:28
Okay, so just to write down what we are going to do.
01:35
Okay, so in part a, we want to find the many two of the e -field and the direction and probably the direction of the electric field.
01:44
Okay, so using kulom's law, here we write our de to be a dq over 4 pi x or not r square.
02:01
So from the diagram our dq will just be lambda dx, lambda is the linear charge density, which is q over l, okay, the x.
02:16
Then our r is going to be x square plus r squared, square root.
02:23
The r is this distance, okay, and the bisymmetry, okay, the horizontal component.
02:38
Cancel out okay so we only need to sum you only need to sum the vertical components okay so the vertical component the e y would be ve on sign data okay and sign data from the diagram from the diagram will be r over square root of x square plus r square okay okay so the electric field at point p is the sum of the y okay so that which is the sum of de side theta okay and so you can you can bring the de expression over here so it's going to be a q or or small q to be exact small q over l x divide by 4 pi epsilon 0 the r square is x squared plus r square and the sign data is r times divide by square root of x square plus r square and then the integration is from negative l over 2 to l over 2 so we put out the constants q r divide by by for pi f dot not l and then you have one over x squared plus r square to the power of three half in the denominator and then the x okay when you reach this step you need to refer to the to the integration table or results at the back of the book so you have q r over the constant space okay and then when you integrate this thing you have one over r square and then x divide by x square plus r square to the power half and then the limit is negative l2 to l2 okay so so you can cancel the r so you have q over for pi epsilon l times r okay and then when you do the limits...