Physics-Electric Fields: Can someone please explain this to me?

In summary, the conversation is about a physics problem that the student is struggling to understand. They are confused about the x and y coordinates and how to calculate the total force. The expert offers guidance and clarifies that the student needs to use the component of F1 and F3 along the diagonal, not the whole force. The student corrects their mistake but is still unsure of the solution. The expert suggests getting common denominators to add the forces, but the student is still struggling to find the correct answer.
  • #1
MitsuShai
159
0
My professor didn't explain this well.

Question: http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q1.jpg


Answer: (part 1) http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q2.jpg
(part 2) http://i324.photobucket.com/albums/k327/ProtoGirlEXE/q3.jpg


I'm completely lost on this one. I don't understand how this problem was solved.

So I'm guessing q4 is the point where you measure the forces from the other 3 points. But I thought that q3 won't have a y value and q1 won't have an x value.
So q2 is only measured by the diagonal right? So it would just be F= k Q^2/(2L^2)---I understand this

why wouldn't q1 and q3 just be F=k Q^2/L^2?

I would like it if someone would explain this whole problem because I feel like I'm completely lost on it.
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
Hi MitsuShai! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)
MitsuShai said:
… But I thought that q3 won't have a y value and q1 won't have an x value.
So q2 is only measured by the diagonal right? So it would just be F= k Q^2/(2L^2)---I understand this

why wouldn't q1 and q3 just be F=k Q^2/L^2?

They are. :smile:

I think you're confused about what the x and y coordinates are.

Your professor has looked at the diagram, and decided that it's obvious that the total force will be along the diagonal …

and so he's decided to make his x coordinate in that direction (instead of along the bottom of the square, as you'd expect).

Does his work make sense now? :smile:​
 
  • #3
tiny-tim said:
Hi MitsuShai! :smile:

(try using the X2 and X2 icons just above the Reply box :wink:)


They are. :smile:

I think you're confused about what the x and y coordinates are.

Your professor has looked at the diagram, and decided that it's obvious that the total force will be along the diagonal …

and so he's decided to make his x coordinate in that direction (instead of along the bottom of the square, as you'd expect).

Does his work make sense now? :smile:​


Oh ok I think I know what you mean. The forces are the diagonal [F= k Q[SUP]2[/SUP]/(2L2)] and the x and y components of the diagonal, which are F=k Q2/L2 each.
And to get the total forces, you have to add up these forces, but the answer is suppose to be [Q2/(8pi*epsilon_0*L2)] (1+2sqrt(2)) and you don't get that with these forces...
 
  • #4
Hi MitsuShai! :smile:

(have a square-root: √ and an epsilon: √ and a pi: π :wink:)
MitsuShai said:
And to get the total forces, you have to add up these forces, but the answer is suppose to be [Q2/(8pi*epsilon_0*L2)] (1+2sqrt(2)) and you don't get that with these forces...

Show us what you get. :smile:

(btw, I've just noticed i should have said "y" not "x" in my last post :rolleyes:)
 
  • #5
tiny-tim said:
Hi MitsuShai! :smile:

(have a square-root: √ and an epsilon: √ and a pi: π :wink:)Show us what you get. :smile:

(btw, I've just noticed i should have said "y" not "x" in my last post :rolleyes:)
F(total)= [ k Q2/(2L2)] + [ k Q2/(L2)] + [ k Q2/(L2)] = [ k Q2/(2L2)] + [2k Q2/(L2)]= [ k Q2/(2L2)] + [4k Q2/(2L2)]= 5k Q2/(2L2)= 3k Q2/(L2)
 
  • #6
Hi MitsuShai! :smile:

(just got up :zzz: …)
MitsuShai said:
F(total)= [ k Q2/(2L2)] + [ k Q2/(L2)] + [ k Q2/(L2)] = [ k Q2/(2L2)] + [2k Q2/(L2)]= [ k Q2/(2L2)] + [4k Q2/(2L2)]= 5k Q2/(2L2)= 3k Q2/(L2)

ah I see …

no, you need to use the component of F1 and F3 along the diagonal, not the whole of F1 and F3 …

try again! :smile:​
 
  • #7
tiny-tim said:
Hi MitsuShai! :smile:

(just got up :zzz: …)ah I see …

no, you need to use the component of F1 and F3 along the diagonal, not the whole of F1 and F3 …

try again! :smile:​
Oh right, I forgot about that part, so
F(total)= [ k Q^2/(2L^2)] + [ k Q^2/(L^2)]sin(45) + [ k Q^2/(L^2)]cos(45)= 2[ k Q^2/(L^2)](1/sqrt(2)) + [ k Q^2/(2L^2)]= [ k Q^2/(2L^2)](1/sqrt(2)) + 4k Q^2/(2L^2)][1/sqrt(2)/(1/sqrt(2)] I'm doing this wrong, somehow... :/
 
Last edited:
  • #8
Hi MitsuShai! :smile:

(please use the X2 icon just above the Reply box :redface:)
MitsuShai said:
2[ k Q^2/(L^2)](1/sqrt(2)) + [ k Q^2/(2L^2)]

that's correct …

i can't see where you've gone wrong after that :confused:
 
  • #9
tiny-tim said:
Hi MitsuShai! :smile:

(please use the X2 icon just above the Reply box :redface:)


that's correct …

i can't see where you've gone wrong after that :confused:


Where do I go from there? I was thinking of adding them and to do that I would need to have common denominators, so I would have to get common denominators and I just noticed that I typed that in wrong... ._.

2[ k Q^2/(L^2)](1/sqrt(2)) + [ k Q^2/(2L^2)]= [ k Q^2/(2L^2)](1/sqrt(2)) + 4k Q^2/(2L^2)](1/sqrt(2))= 5k Q^2/(4L^2), which isn't right
 

1. What is an electric field?

An electric field is a region of space surrounding an electrically charged particle or object where other charged particles will experience a force.

2. How is an electric field created?

An electric field is created when a charged particle or object exerts a force on another charged particle or object. This can happen through direct contact, induction, or by applying an external electric field.

3. What is the difference between an electric field and an electric potential?

An electric field is a vector quantity that represents the direction and strength of the force experienced by a charged particle. Electric potential, on the other hand, is a scalar quantity that represents the potential energy of a charged particle at a specific point in the electric field.

4. How is the strength of an electric field measured?

The strength of an electric field is measured in volts per meter (V/m) or newtons per coulomb (N/C). This represents the amount of force experienced by a charged particle per unit charge.

5. Can you give an example of how electric fields are used in everyday life?

Electric fields are used in many everyday devices, such as televisions, computers, and cell phones. They are also used in power lines to transport electricity from power plants to homes and businesses.

Similar threads

Forces and torque on a cable/beam
    • Introductory Physics Homework Help
Replies
7
Views
4K
Solving Helmholtz Coils: Diagram, Magnetic Field & Current
    • Introductory Physics Homework Help
Replies
1
Views
1K
Can someone grade my practice test?
    • Introductory Physics Homework Help
Replies
5
Views
1K
Calculating Two Unknown Charges from a Tripole
    • Introductory Physics Homework Help
Replies
3
Views
3K
How does capacitance affect voltage in a capacitor?
    • Introductory Physics Homework Help
Replies
4
Views
9K
Direction fields: Don't understand them.
    • Calculus and Beyond Homework Help
Replies
15
Views
3K
Flow of electricity and resistivity (ohms law)
    • Introductory Physics Homework Help
Replies
4
Views
5K
What Mistakes Did I Make in These Trigonometric Integral Problems?
    • Calculus and Beyond Homework Help
Replies
16
Views
3K
Find the electric field at the point 20cm above the centre
    • Introductory Physics Homework Help
Replies
1
Views
1K
Electrostatic Force and Field between two charges
    • Introductory Physics Homework Help
Replies
5
Views
1K

Hot Threads

Recent Insights

Back
Top