Calculating total Coulomb force vector ?

Click For Summary
The discussion focuses on calculating the total Coulomb force vector acting on a positive charge in a configuration with three negative charges arranged in a square. The user attempts to break down the forces into components, specifically F1, F2, and F3, using the vector form of Coulomb's force. There is some confusion regarding the correct representation of unit vectors and the calculations for the diagonal force vector. The response clarifies that it's more efficient to remain in Cartesian coordinates and suggests directly summing the components of the forces in the x and y directions. The overall approach is deemed acceptable, with emphasis on simplifying the calculations without switching coordinate systems.
starstruck_
Messages
185
Reaction score
8

Homework Statement


Consider a configuration consisting one +q charge ( upper right) and three −q charges, arranged in a square.

Side lengths = d.

Calculate the total F force vector acting on charge +q.

Homework Equations


Vector form of culomb’s force
F=( kq1q2/r^2) rhat

(rhat for unit vector - I’m on my phone so I can’t really tyupe it out properly, sorry)

The Attempt at a Solution


Split into 3 Force vector (between +q and each charge ). F1 is horizontal, F2 is the diagonal force vector, F3 is the vertical force vector.

F1 = (-ke^2/r^2)rhat

r= d
rhat= (ihat)
Or is it rhat= d( ihat )?

Anyways,
F1= ((-ke^2)/d^2))(ihat)

F2 = (-ke^2/r^2)(rhat)
r = sqrt(d^2+d^2)
r= sqrt(2)d
rhat = r vector/|r vector|
Assuming tail at origin
r vector = <d,d>
rhat= d(ihat)+d(jhat)/(sqrt(2)d)

F2= (-ke^2/(2d^2))(d(ihat)/sqrt(2)d)+ (-ke^2/(2d^2))(d(jhat)/sqrt(2)d)

F2 =(-ke^2/(2sqrt(2)d^3))(d(ihat))+(-ke^2/(2sqrt(2)d^3))(d(jhat))

F3 = (-ke^2/r^2)(rhat)

r= d
rhat= jhat or is it rhat = d(jhat)?

F3= (-ke^2/d^2)(jhat)

And then I would just add all of the components together.

I’m just wondering if I did this rhat business correctly ?

It said to use the vector form of the Coulomb force, so I tried- I’m not used to working with forces in this manner.

((Also sorry if it’s really hard to understand what I did, I can try to write it down and post a picture of my work if possible)).
 
Physics news on Phys.org
Your basic approach is OK.

It's convenient in this forum to use bold letters to convey vectors.
So r_hat = r, i_hat = i, j_hat = j, r =r cosθ i + r sinθ j and F = Fx i + Fy j.

When you introduce r you are doing coordinate system switching between cartesian (x,y) and polar (r,θ) coordinates. But this is not necessary. You can just stick with cartesian. So for example F2 = kq/d2 i + kq/d2 j and so on for F1 and F3. Then just add all the x and y components separately to get the net force in the i and j directions. Note that you don't calculate r2 =2d2separately. At the end you can still compute r (and θ) if you want.
 
  • Like
Likes starstruck_
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 6 ·
Replies
6
Views
1K
Replies
17
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
3
Views
1K
Replies
3
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
1
Views
3K