A point charge q1 60 nc is at the point 030 m 000 m a charge...

A point charge Q1 = +6.0 nC is at the point (0.30 m, 0.00 m); a charge Q2 = -1.0 nC is at (0.00 m, 0.10 m), and a charge Q3 = +5.0 nC is at (0.00 m, 0.00 m). What are the magnitude and direction of the net force on the +5.0 nC charge due to the other two charges?

Transcript

00:01 Hello everyone let's let's seek to solve this question so let's draw the coordinate system first so this is our y -axis and this is our x -axis here all right so we have our origin somewhere in the middle in the intersection point so this is the origin here all right now a charge q1 so here is discharge q1 which is whose magnitude is six nanoculums and it is at a distance of 0 .3 .0 from the origin.

00:38 Another charge q2 so another charge q2 is here and its magnitude is given us to be negative of 1 nanoculums.

00:49 All right and it is at a distance of so the coordinates are 0 .0 .1 from origin and finally a charge q3 is at the origin at 0 .0 and the magnitude of charge q3 is 5 nanoculum so we need to find the intensity the magnitude and direction of the force which is been applied on 5 nanoculum charge all right so let us first try to find the force.

01:26 So the electrostatic force f is given as k, q1, q2, divided by the square of distance between them, where k is a constant, which value is 9 multiplied by 10 to the power 9.

01:42 All right? so let's find the force as the charge q1 and q3 both are positive, both are like charges.

01:51 So q1 will repel charge q3 in this direction.

01:59 All right.

02:00 So this is the force being applied on charge q3.

02:04 So there will be a force of repulsion.

02:06 This let's write is as f1.

02:11 So f13 is the force applied on charge q3 by q1.

02:17 So this will simply be equal to k multiplied by q1, q3.

02:23 And the distance between them.

02:24 So the distance between them is 0 .3.

02:29 So 0 .3 square.

02:32 So let's substitute all the values to find f1 3...