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Electric Charges And Fields

Question
CBSEENPH12037283

A uniform line charge of linear charge density λ coulombs per metre exists along the x-axis from x = – a to x = + a. Find the electric field intensity E at a point P at a distance r along the perpendicular bisector.


Solution

In all the problems, which involve distribution of charge, we choose an element of charge dq to find the element of the field dE produced at the given location. Then we sum all such dEs to find the total field E at that location.
We must note the symmetry of the situation. For each element dq located at positive x-axis. There is a similar dq located at the same negative value of x. The dEx produced by one dq is cancelled by the dEx in the opposite direction due to the other dq. Hence, all the dExcomponents add to zero.
So, we need to sum only the dEy components, a scalar sum since they all point in the same direction.
The element charge is dq = λ dx


The integral on the right hand side can be evaluated by substituting x = r tan α and dx = r sec2 α.


                                              [sin α = xx2+r2]