Magnitude of Force of two point charges on a third (electrostatics)

[ctallarico20]

Two point charges are placed on the x-axis as follows: charge q1 = 4.01nC is located at x= 0.201m , and charge q2 = 5.00nC is at x= -0.301m.

What is the magnitude of the total force exerted by these two charges on a negative point charge q3 = -6.03nC that is placed at the origin?

I know I need to use:
F = (k * q_1 * q_2)/d^2, where k = 9E9


I've tried taking the sum of magnitudes of F(1 on 3) and F(2 on 3), but have been incorrect. I set it up like below:

F(1 on 3) = (k * 4.01E-9 * -6.03E-9) / 0.201^2 = -0.000005 N

F(2 on 3) = (k * 5.00E-9 * -6.03E-9) / 0.301^2 = -0.000003 N


This would give the answer as 0.000008 N, or more precisely 8.37E-6 N. Could anyone tell me where I'm going wrong?

 

[rude man]

check the sign of F(2 on 3). Which direction is it?

 

[ctallarico20]

Yeah, it should be positive, my bad. However I've also tried 0.000002 as an answer and that was incorrect:/

 

[gneill]

Lose the fixed point notation and stick to scientific notation so that significant figures are clear. What's the value of the net force to three significant figures?

 

[ctallarico20]

F(1 on 3) = -5.387E-6
F(2 on 3) = -2.995E-6

This is a magnitude of 8.37E-6 N, which was incorrect

 

[gneill]

F(1 on 3) = -5.387E-6
F(2 on 3) = -2.995E-6

This is a magnitude of 8.37E-6 N, which was incorrect

The first force is wrong (direction).
The second force is okay.

 

[ctallarico20]

The magnitude appears to be correct though, right? For some reason it's not

 

[rude man]

The magnitude appears to be correct though, right? For some reason it's not

Probably because I mixed you up. I suggested F(2 on 3 ) should be +, but it's -. Similarly, F(1 on 3) should be +, not -. But the magnitude of the difference is the same.

Do it gneill's way which is right.

 

[ctallarico20]

Probably because I mixed you up. I suggested F(2 on 3 ) should be +, but it's -. Similarly, F(1 on 3) should be +, not -. But the magnitude of the difference is the same.

Do it gneill's way which is right.

In that case, F(1 on 3) is = 5.387E-6
F(2 on 3) is = -2.995E-6

Magnitude still appears to be 8.37E-6 N like you said. Am I using the wrong equation or something? This doesn't make sense to me

 

[gneill]

Add the forces, then take the magnitude.

 

[ctallarico20]

Ohhh I gotcha, so 5.387E-6 + (-2.995E-6), which = 2.392E-6. So the magnitude is 2.392E-6N?

 

[gneill]

Yup. Round to 3 significant figures for your final result.

 

[ctallarico20]

Perfect. Thank you all so, so much