Q1 = +2.60 x 10^-6 C, q2 = +3.75 x 10^-6 C, and q3 = +1.30 x...

q1 = +2.60 x 10^-6 C, q2 = +3.75 x 10^-6 C, and q3 = +1.30 x 10^-6 C. Find the y-component of the net force on q1. Include the correct + or - sign to indicate direction. q1 0.283 m 0.200 m 45.0° q2 0.200 m q3 (Make sure you know the direction of each force!) y-component (N)

Transcript

00:01 Hello, in the question we have to find out the y -component of the net force on this charge q1.

00:08 So, and provided we have given some information that charge q1 is 2 .6 into 10 raised to minus 6 coulomb, then q2 is 3 .75 into 10 raised to minus 6 coulomb and q3 is 1 .3 into 10 raised to minus 6 coulombs and they have given this distance, the distance between q2 and q3 is 0 .2, then the distance between q3 and q1 is 0 .2 and the distance between q2 to q1 is 0 .285 meters.

00:41 So, we have to find out only the y -component.

00:45 So, let us first find out the force on q1 due to this charge q2.

00:53 So, force on q1 due to this charge q2, so it will be f on charge 1 due to charge q2, so that will be kq1q2 divided by r square, where r is the distance between these two charges which is 0 .285.

01:20 Now see this force, this charge is positive, this charge is also positive, so the force will be repulsive and it will be directed in this, so this is the force f12.

01:33 So, let us first calculate the magnitude.

01:37 So, it will be k is 9 into 10 raised to 9 into q1 is 2 .60 into 10 raised to minus 6 into 3 .75 into 10 raised to minus 6 divided by this r square, r is 0 .283 square, it is 0 .285, sorry, so it is 0 .285 square.

02:18 So, if we calculate this, we will get this force as equal to, so f12 will be equal to 1 .0956 newtons or 0 .19 if we just approximate it.

02:40 So, this will be the force f12.

02:42 Now coming over here back, so this angle is 45, so if i draw this line like this, so if this angle is 45, then this angle will also be 45.

02:56 So, then this component, this component of f12 will be f12 cosine of theta, that is the x component and the y component will be like this, which will be f12 sine of 45.

03:14 So, we are interested only in the y component.

03:18 So, f12 y will be equal to f12 sine of 45.

03:27 So, f12 is 1 .0956 divided by sine of 45 is 1 by root 2...

Question

q1 = +2.60 x 10^-6 C, q2 = +3.75 x 10^-6 C, and q3 = +1.30 x 10^-6 C. Find the y-component of the net force on q1. Include the correct + or - sign to indicate direction. q1 0.283 m 0.200 m 45.0° q2 0.200 m q3 (Make sure you know the direction of each force!) y-component (N)

Please give Ace some feedback