00:01
In this problem, we have been given the three point charges and we have to first determine the electric field in terms of vectors at the origin because of 6 nanoculum and minus 3 nanoculum charge.
00:16
So to compute the electric field, we use the expression e is equal to kq by r square.
00:22
R here is the distance of the point with respect to the charge.
00:25
And we know the direction of electric field that's positive to the negative.
00:29
So here we observe at the origin, the electric field because of 6 nanoculum charge that will be along the negative x direction.
00:38
And let's represent this as e6.
00:42
And the electric field because of minus 3 nanoculum charge that will be along the negative y direction because the electric field is from positive to the negative.
00:52
And let's represent this using e3.
00:54
And now we can say that the net electric field, that will be.
01:00
Minus e6 i cap minus e3 j cap and let's put the values of e3 and e6 so e6 that will be k times the charge that six times ten raised to minus nine over point three square i cap minus e3 which will be k times three into ten raise to minus nine over point one square j cap and now when we put the value of k here as nine and simplify the this expression we're going to get here the result coming out to be minus 600 i cap and this will be here again we put the value of k as nine and we get that on simplification this comes out to be 2700 and that will be in j cap so this is the net electric field at the origin because of 6 nanoculum charge and minus 3 nanoculum charge.
02:07
And now we have to determine the force that the 5 nanoculum charge placed at this origin experiences.
02:15
So to compute the force, let's use the idea that force is the electric field per unit charge.
02:22
And now in order to compute, it's a posit here because we are representing electric field using e and force using f...