Getting direction of force of two point charges on a third

Click For Summary
The discussion revolves around determining the direction of the force on a -10 nC charge due to two other point charges. The user initially calculated the force components but struggled with the correct angle measurement from the positive x-axis. They realized they had mixed up the signs and axes in their calculations, which led to the incorrect angle despite having the correct force magnitude. After correcting the sign errors and adjusting the input for the arctangent function, they were able to find the correct answer. The focus on accurate component analysis and angle calculation is crucial in solving such electrostatic 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).
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 10 ·
Replies
10
Views
7K
  • · Replies 23 ·
Replies
23
Views
1K
  • · Replies 21 ·
Replies
21
Views
2K
Replies
1
Views
818
  • · Replies 7 ·
Replies
7
Views
796
Replies
3
Views
516
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
4
Views
2K