I can graphically visualize this, but algebraically, how do you prove it? 11 a)

Question

I can graphically visualize this, but algebraically, how do you prove it? 11 a) Three point charges are positioned as follows: q1 is at (0 m, 0 m), q2 is at (1.2 m, 0 m), and q3 is at (1.2 m, 1.6 m). If q1 = 1.00 C, q2 = 2.00 C, and q3 = 3.00 C, in what direction (ccw from the x-axis) is the force on q2 ? b) What is the magnitude of the force on q2 ? I can graphically visualize this, but algebraically, how do you prove it? 11 a) Three point charges are positioned as follows: q1 is at (0 m, 0 m), q2 is at (1.2 m, 0 m), and q3 is at (1.2 m, 1.6 m). If q1 = 1.00 C, q2 = 2.00 C, and q3 = 3.00 C, in what direction (ccw from the x-axis) is the force on q2 ? b) What is the magnitude of the force on q2 ?

Explanation / Answer

I can graphically visualize this, but algebraically, how do you prove it?

11 a) Three point charges are positioned as follows: q1 is at (0 m, 0 m), q2 is at (1.2 m, 0 m), and q3 is at (1.2 m, 1.6 m). If q1 = 1.00 C, q2 = 2.00 C, and q3 = 3.00 C, in what direction (ccw from the x-axis) is the force on q2 ?

Use F=kq1q2/r^2 to determine each force on q2:

1) F12=((9x10^9)(1x10^-6)(2x10^-6))/(1.2^2)=1.2x10^-2N

This force is entirely parallel to q2. Also, both charges are positive, so q1 will exert a repulsive force on q2 in the positve x direction of the axis.

2) F23=((9x10^9)(3x10^-6)(2x10^-6))/(1.6^2)=2x10^-2N

This force is perpendicular to q2, and also exerts a repulsive force, though this force pushes down on q2 in the negative y direction.

So we have one force pushing the south and another force pushing east, or a total force pushing q2 in the southeast direction. We can be a little more specific about this:

---find the combined length of q12 and q32. You'll realize that we have a right triangle and can simply use pythagorean's theorem to find the hypotenuse (which will be the combination of the two x and y vectors):

(a^2 + b^2)=c
(1.2^2 + 1.6^2)=c c=2m To find the angle: sin=opposite/hypotenuse=1.6/2=.8 --> asin(.8)=53.13: =53.13 in south east direction.

b) What is the magnitude of the force on q2 ? You'll use the pythagorean theorem again, only this time will the forces we found earlier. ((1.2x10^-2)^2) + (2x10^-2)^2)=c c=2.3x10^-2N So the combination of q3 and q1's force on q2 equals 2.3x10^-2N at a 53.13 angle in the southeast direction.

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