00:01
Hi, here in this given problem first of all we draw a cartesian plane.
00:06
Cartesian plane having x axis horizontal, y axis vertical.
00:13
This is the origin.
00:15
One of the charge is put at the origin and that charge is plus 2 .50 micro coulomb.
00:23
Another charge is put at the positive x axis.
00:26
Suppose at a point 8 the charge is minus 3 .50 micro coulomb.
00:34
The gap between these two charges means position of second charge at the x axis.
00:38
This is 0 .600 meter.
00:43
Ok, now net force on any charge on a charge plus q.
00:58
Will be 0 at a point where net electric field will be 0.
01:09
And due to a group of two unlike charges, unlike charges means one positive and another negative.
01:30
Electric field is 0 outside the charge which is smaller in magnitude.
01:50
Your electric field can never be 0 between the two unlike charges.
01:56
It is 0 outside that point, outside that charge which is smaller in magnitude.
02:04
And here that is 2 .50 micro coulomb which is smaller in magnitude.
02:10
So suppose electric field is 0 at a point b which is at a distance x from the origin.
02:19
So at this point b there will be two electric fields.
02:25
Electric field at b due to the charge put at o.
02:31
That should be equal to electric field at b due to the charge which is put at a.
02:36
And using the expression for electric field for ebo that is k into q...