Saved Bookmarks
| 1. |
Calculate the electric field at point P,Q for the following two cases as shown in the figure (a) A positive point charge +1 muCis placed at the origin (b) A negative point charge -2 muCis placed at the origin |
|
Answer» Solution :The magnitude of the electric field at point P is `E_(p)=(1)/(4piepsilon_(0))=(q)/(r^(2))=(9xx10^(9)xx1xx10^(-6))/(4)=2.25xx10^(3)NC^(-1)` Since the source charge is positive the electric field points AWAY from the charge .So electric field at the point P is given by `vecE_(p)=2.25xx10^(3)NC^(-1)hatj` For the point Q `|vecE_(Q)|=(9xx10^(9)xx1xx10^(-6))/(16)=0.56xx10^(3)NC^(-1)` Hence `vecE_(Q)=0.56xx10^(3)hatj` Case (b) : The magnitude of the electric field at point P `|vecE_(p)|=(kq)/(r^(2))=(1)/(4piepsilon_(0))(q)/(r^(2))=(9xx10^(9)xx2xx10^(-6))/(4)` `=4.5xx10^(3)NC^(-1)`  Since the source charge is negative the electric field points towards the charge . So the electric field at the point P is given by `vecE_(p)=-4.5xx10^(3)hatiNC^(-1)` For the point Q, `|vecE_(Q)|=(9xx10^(9)xx2xx10^(-6))/(36)` `=0.5xx10^(3)NC^(-1)` `vecE=0.56xx10^(3)NC^(1)` At the point Q the electric field is directed along the positive x -axis . \ |
|