Getting direction of force of two point charges on a third

Click For Summary
SUMMARY

The discussion focuses on determining the direction of the force acting on a -10 nC charge due to two other point charges, specifically a +15 nC charge and another -5 nC charge. The calculations involved using Coulomb's Law, represented by the equation F = k(Qq/r²), to find the components of the forces F1 and F2. The user initially miscalculated the angle due to sign errors in the force components but corrected this by switching the x and y components, ultimately arriving at the correct angle of 160° measured clockwise from the positive x-axis.

PREREQUISITES
  • Coulomb's Law for electrostatic force calculations
  • Vector decomposition in two dimensions
  • Trigonometric functions, specifically tangent and inverse tangent
  • Understanding of significant figures in scientific calculations
NEXT STEPS
  • Review vector addition and decomposition in physics
  • Study Coulomb's Law applications in electrostatics
  • Practice calculating angles using inverse trigonometric functions
  • Explore common mistakes in vector calculations and how to avoid them
USEFUL FOR

Students in physics, particularly those studying electrostatics, as well as educators looking for examples of common calculation errors in force direction problems.

cmkluza
Messages
118
Reaction score
1

Homework Statement


What is the direction of the force F on the -10 nC charge in the figure? Give your answer as an angle measured cw from the +x-axis.
Express your answer using two significant figures.
26.P39.jpg

Homework Equations


##F = k\frac{Qq}{r^2}##

The Attempt at a Solution


I started by getting the components of the net force. The force between the two negative charges will be ##F_1## and the one between the -10 and +15 charges will be ##F_2##.
##F_1 = <0, -y> \longrightarrow F_{1x} = k\frac{(5\times10^{-9})(10\times10^{-9})}{0.01^2}##
##F_1 = <0, -0.004495>##
##F_2 = <-x, y> \longrightarrow F_{2t} = k\frac{(15\times10^{-9})(10\times10^{-9})}{(0.03^2 + 0.01^2)}##
##\theta = \tan^{-1}(\frac{1}{3}) \longrightarrow F_2 = <-\cos(\theta)\bullet F_{2t}, \sin(\theta)\bullet F_{2t}>##
##F_t = <-0.0012793, -0.0040685>##
All of this math appears to be correct, because when I calculate the net force (##F_t = \sqrt{x^2 + y^2} \approx 4.3\times10^{-3}##) I get the right answer.
So, now I just need to find the angle, in degrees, clockwise from the positive x-axis, but whatever I'm doing appears to be wrong. Given these components, the vector should be in the third quadrant. So, I tried calculating the inverse tangent of the y over the x components of the total force (##\tan^{-1}(\frac{y}{x}) \approx 17.45##°), and I should be getting the right angle (I've even drawn it out and checked multiple times) if I subtract this answer from 180°, giving 162.55°, which should be 160° to 2 significant figures. But this is incorrect. What am I missing?
 
Last edited:
Physics news on Phys.org
cmkluza said:

The Attempt at a Solution


I started by getting the components of the net force. The force between the two negative charges will be ##F_1## and the one between the -10 and +15 charges will be ##F_2##.
##F_1 = <-x, 0> \longrightarrow F_{1x} = k\frac{(5\times10^{-9})(10\times10^{-9})}{0.01^2}##
##F_1 = <-0.004495, 0>##
I haven't checked your entire calculation. But, does F1 point in the negative x direction?
 
TSny said:
I haven't checked your entire calculation. But, does F1 point in the negative x direction?
Yeah, I mixed up a lot of signs and axes in my OP, but it looks like I did so in my work as well - I had the x- and y-components for the final force switched up, hence why I got the right force magnitude but not direction. Thanks for pointing it out! That let me get the right answer (just had to switch the numbers I was putting in the arctan).
 

Similar threads

  • · Replies 10 ·
Replies
10
Views
7K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 21 ·
Replies
21
Views
3K
Replies
1
Views
896
  • · Replies 7 ·
Replies
7
Views
893
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K