Find the magnitude of the electric force from 3 charges at vertices of a cube

In summary: N.Yes, the distance between B and P is greater than the distance between A and P. I took ##k = 9\times 10^9 Nm^2/C^2##
  • #1
paulimerci
287
47
Homework Statement
Question is posted below.
Relevant Equations
Coulomb law F_e = kq1q2/r^2
There are three charges with +1 μC and −1 μC, are placed at the opposite corners of a cube with edges of length 1 cm, and the distance from P to B is 1cm 2. I labeled them as A, P, and B, which is shown in the diagram below. Since we need to find the magnitude of the charge at point P and the charge at point P is +1 μC, it exerts an attractive force on point B and a repulsive force on point A. I've indicated the FBD below. Using Coulomb's law, we get
Fe=kq1q2r2
∑Fx=FBPcos45+FAP
∑Fx=91

∑Fy=FBPsin45
∑Fy=45.5N
Using Pythagorean we get,
Resultant magnitude of electric force = 101.7N
I did something wrong here. Can anyone point out where I made the mistake?
Thank you!
 

Attachments

  • Screenshot 2023-05-17 at 6.27.29 PM.png
    Screenshot 2023-05-17 at 6.27.29 PM.png
    8.4 KB · Views: 79
  • Screenshot 2023-05-17 at 6.18.31 PM.png
    Screenshot 2023-05-17 at 6.18.31 PM.png
    16.9 KB · Views: 90
Last edited:
Physics news on Phys.org
  • #2
Are you sure that force ##\mathbf{F}_{BP}## has a nonzero x-component?
 
  • #3
TSny said:
Are you sure that force ##\mathbf{F}_{BP}## has a nonzero x-component?
Does that mean F_BP cancels with F_AP since they are in opposite directions?
 
  • #4
TSny said:
Are you sure that force ##\mathbf{F}_{BP}## has a nonzero x-component?
I got the answer, and I edited my answer in post #1.
 
  • #5
paulimerci said:
Fe=kq1q2r2
∑Fx=FBPcos45+FAP
∑Fx=91
The equation ##\sum F_x = F_{BP}\cos 45^o + F_{AP}## is not correct.
Can you see that the vector ##\mathbf{F}_{BP}## is perpendicular to the vector ##\mathbf{F}_{AP}##?

paulimerci said:
∑Fy=FBPsin45
∑Fy=45.5N
##F_{BP}\sin 45^o \neq 45.5 N##. What value did you get for ##F_{BP}##?

Does the vector ##\mathbf{F}_{BP}## have a z-component as well as a y-component?
In your diagram, you didn't indicate the directions of the x, y, and z axes.
 
Last edited:
  • Like
Likes scottdave and MatinSAR
  • #6
TSny said:
The equation ##\sum F_x = F_{BP}\cos 45^o + F_{AP}## is not correct.
Can you see that the vector ##\mathbf{F}_{BP}## is perpendicular to the vector ##\mathbf{F}_{AP}##.##F_{BP}\sin 45^o \neq 45.5 N##. What value did you get for ##F_{BP}##?

Does the vector ##\mathbf{F}_{BP}## have a z-component as well as a y-component?
In your diagram, you didn't indicate the directions of the x, y, and z axes.
Isn’t the ##F_{BP} sin45## perpendicular to the ##F_{AP}##?
F_BP sin 45 = 31.8N
 
  • #7
The key is to realize that the vector ##\mathbf{F}_{BP}## is perpendicular ##\mathbf{F}_{AP}##.

1684380242711.png

If you see that the two vectors are perpendicular, then it should be easy to find the magnitude of their sum. You don't need to consider x, y, or z components.
 
  • Like
Likes scottdave, paulimerci, MatinSAR and 1 other person
  • #8
TSny said:
The key is to realize that the vector ##\mathbf{F}_{BP}## is perpendicular ##\mathbf{F}_{AP}##.

View attachment 326712
If you see that the two vectors are perpendicular, then it should be easy to find the magnitude of their sum. You don't need to consider x, y, or z components.
Thanks for the diagram. I can't visualize F_BP perpendicular to F_AP.
 
  • #9
paulimerci said:
Thanks for the diagram. I can't visualize F_BP perpendicular to F_AP.
##\mathbf{F}_{BP}## lies in the face of the cube that is shaded yellow.

The edge AP of the cube is perpendicular to the yellow face.

Imagine looking at the cube so that you are looking directly at the yellow face.
1684381043625.png

Imagine what the direction of ##\mathbf{F}_{AP}## would be in this picture.
 
  • Like
Likes scottdave, paulimerci, MatinSAR and 1 other person
  • #10
TSny said:
##\mathbf{F}_{BP}## lies in the face of the cube that is shaded yellow.

The edge AP of the cube is perpendicular to the yellow face.

Imagine looking at the cube so that you are looking directly at the yellow face.
View attachment 326713
Imagine what the direction of ##\mathbf{F}_{AP}## would be in this picture.
I think I understood what you said. So the magnitude of the electric force is 128N?
 
  • #11
paulimerci said:
I think I understood what you said. So the magnitude of the electric force is 128N?
Please show the calculation that gave you this result.

What value did you get for ##F_{AP}##?

What value did you get for ##F_{BP}##?
 
  • #12
TSny said:
Please show the calculation that gave you this result.

What value did you get for ##F_{AP}##?

What value did you get for ##F_{BP}##?
Ok.
##F_{AP} = 91N##
##F_{BP} = 91N##
Using Pythagorean, the net magnitude of electric force at point p is 128N.
 
  • #13
paulimerci said:
##F_{AP} = 91N##
I'm not sure how you got 91 N rather than 90 N. The value of Coulomb's constant is ##k = 8.99 \times 10^9## Nm2/C2.

paulimerci said:
##F_{BP} = 91N##
The distance between B and P is greater than the distance between A and P. So, shouldn't ##F_{BP}## be less than ##F_{AP}##?
 
  • Like
Likes scottdave and paulimerci
  • #14
TSny said:
I'm not sure how you got 91 N rather than 90 N. The value of Coulomb's constant is ##k = 8.99 \times 10^9## Nm2/C2.The distance between B and P is greater than the distance between A and P. So, shouldn't ##F_{BP}## be less than ##F_{AP}##?
Yes, the distance between B and P is greater than the distance between A and P. I took ##k = 9\times 10^9 Nm^2/C^2##
I did something wrong in calculating the distance.
##F_{BP} = 45.5 N ## and ##F_{AP} = 91N##.
Thereby the magnitude of electric force at point P is 102N.
 
  • #15
paulimerci said:
Yes, the distance between B and P is greater than the distance between A and P. I took ##k = 9\times 10^9 Nm^2/C^2##
I did something wrong in calculating the distance.
##F_{BP} = 45.5 N ## and ##F_{AP} = 91N##.
Thereby the magnitude of electric force at point P is 102N.
Using ##\displaystyle k = 9\times 10^9 \, \rm{Nm^2/C^2}## rather than ##\displaystyle 8.99\times 10^9 \, \rm{Nm^2/C^2}## will give ##F_{AP} = 90\, \rm N##
 
  • #16
SammyS said:
Using ##\displaystyle k = 9\times 10^9 \, \rm{Nm^2/C^2}## rather than ##\displaystyle 8.99\times 10^9 \, \rm{Nm^2/C^2}## will give ##F_{AP} = 90\, \rm N##
I'm sorry I used ## k= 9.1 \times 10^9 Nm^2/C^2## that's how I got the answer for ##F_{AP} = 91N ##
 
  • #17
paulimerci said:
##F_{BP} = 45.5 N ## and ##F_{AP} = 91N##.
Thereby the magnitude of electric force at point P is 102N.
OK. You have the solution. Using k = 9.0 x 109 Nm2/C2 would give 101 N. But the data in the problem is stated to only one significant figure! Your answer is close enough to pick out the correct answer in the multiple choice.
 
  • Like
Likes scottdave, MatinSAR and paulimerci
  • #18
TSny said:
OK. You have the solution. Using k = 9.0 x 109 Nm2/C2 would give 101 N. But the data in the problem is stated to only one significant figure! Your answer is close enough to pick out the correct answer in the multiple choice.
The answer is close to 100 N. Thank you @TSny for your great help!
 
  • Like
Likes berkeman, TSny and MatinSAR

FAQ: Find the magnitude of the electric force from 3 charges at vertices of a cube

How do you determine the positions of the charges in a cube?

To determine the positions of the charges in a cube, you need to identify the coordinates of each vertex. For a cube with side length \(a\), the vertices can be represented as (0,0,0), (a,0,0), (0,a,0), (0,0,a), (a,a,0), (a,0,a), (0,a,a), and (a,a,a).

What is Coulomb's Law and how is it applied here?

Coulomb's Law states that the electric force \(F\) between two point charges \(q_1\) and \(q_2\) separated by a distance \(r\) is given by \(F = k_e \frac{q_1 q_2}{r^2}\), where \(k_e\) is Coulomb's constant. In the context of charges at the vertices of a cube, this law is applied to calculate the force between each pair of charges, considering their respective distances.

How do you calculate the distance between charges located at different vertices of a cube?

The distance between charges at different vertices of a cube can be calculated using the Euclidean distance formula. For two vertices with coordinates \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), the distance \(r\) is given by \(r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\).

How do you find the net electric force on a charge due to multiple other charges?

To find the net electric force on a charge due to multiple other charges, you need to calculate the vector sum of the forces exerted by each individual charge. This involves determining the magnitude and direction of each force using Coulomb's Law and then summing the vector components (x, y, and z) of these forces.

Can you provide a step-by-step example of calculating the electric force in a specific cube configuration?

Sure! Consider a cube with side length \(a\) and charges \(q\) at vertices (0,0,0), (a,0,0), and (0,a,0). First, calculate the distances between the charges. For example, the distance between (0,0,0) and (a,0,0) is \(a\), and between (0,0,0) and (0,a,0) is also \(a\). Using Coulomb's Law,

Back
Top