Electrostatic Physics Problem, Can't Figure Out The Direction Of The Resultant Force?

Question

The Question says:

A charge Q1 = -16 x 10 -6 coulomb is fixed on the x-axis at +4.0 m and a charge of Q2 = +9.0 x 10 -6 coulomb is fixed on the y-axis at +3.0 meters as shown above. A charge of Q3 = -4.0 x 10 -6 coulomb is brought from a very far point by an external force and placed at the origin.

1. Draw the forces F1,F2 and the resultant force on an axis like the one below and calculate the magnitude of the resultant force and its direction.

I'm having trouble finding the direction of the force, which comes out to me to be 45 degrees, but that does not make sense for this situation--I suspect it should be between 90 degrees and 180 degrees. Help me out if you can, thanks!

Answer 1

Hi Ashley!

Let's take a look at the problem. As you know, the equation for the force between two charged particles is:

F=kq₁q₂/r²

In the x-direction, we are looking at the force that Charge 1 puts on Charge 3. When dealing with this equation, we always assume that the charges will repel, and the resulting sign at the end will determine if our assumption is correct, or if the assumption was opposite:

Q₁ is 4 meters away in the positive x-direction and has a charge of -16x10^-6 Coulombs.

F₁₃ = kq₁q₃/r² = (9x10⁹)(-16x10^-6)(-4.0x10^-6)/16 = 0.036N

Since we have a positive result, the assumption of repulsion was correct, and Q₃ will experience a 0.036N force in the negative x-direction due to Q₁.

Now in the Y-Direction, Q₂ is 3 meters away and has a charge of 9.0x10^-6. Again, our assumption will be repulsion:

F₂₃ = kq₂q₃/r² = (9x10⁹)(9.0x10^-6)(-4.0x10^-6)/16 = -0.036N

Since we have a negative result, the assumption of repulsion was incorrect, and Q₃ will experience a 0.036N force in the positive y-direction due to Q₂.

So we have our two vectors: 0.036N in the positive Y direction, and 0.036 in the negative X-Direction. The magnitude of this force is just pythagorean's theorem:

F²=F_x²+F_y², F² = (-0.036)²+(-0.036)²= 0.002529, F = 0.0509N

For direction, we can do this intuitively - we know that if the magnitudes are the same, the direction will be some odd multiple of 45 degrees (45, 135, 225, or 315). The Y-Direction force will pull us into either Quadrants 1 and 2, while the X-Direction force will pull us into either Quadrants 2 and 3. Both of these pulls have Quadrant 2 in common, so the angle is 135 degrees.

The mathematical way to to use the inverse tangent function on the components we calculated:

tan(theta) = Y/X

One thing to note about the inverse tangent function is that, when X is negative, we must add 180 degrees to it. This is because the tangent function is only valid for positive X values.

So really, our angle is:

theta = tan^-1(y/x) + 180 = tan^-1(0.036/-0.036) +180 = -45 + 180 = 135 degrees

Thus confirming our geometric approach.

Hope this helps!

Christopher