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Electrostatic force between more than 2 charges

  • #1
hi
i dont know if this has been posted or not but i am sort fo stuck on this one

What is the force F_vec on the 1 nC charge at the bottom? View Figure
Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to three significant figures.

the figure looks like this knight_Figure_25_46.jpg

attempt to this problem:

i found out the forces individually and added them together .. but it didnt work

x-component:
6nC = 0
2nC = 5.09 x 10[tex]^{-5}[/tex] each

y-component:
6nC = 2.16 x 10[tex]^{-5}[/tex]
2nc = 5.09 x 10[tex]^{-5}[/tex] each

when i add them: 1.02 x 10 [tex]^{-5}[/tex] [tex]\hat{x}[/tex] , 7.76 x 10[tex]^{-5}[/tex] [tex]\hat{y}[/tex]

but that was wrong.. i also tried the x-comp to be 0 .. but that was wrong too

thanx in advance for the help
 

Answers and Replies

  • #2
Doc Al
Mentor
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How does the x-component of the force exerted by one 2 nC charge compare to that of the other?
 
  • #3
How does the x-component of the force exerted by one 2 nC charge compare to that of the other?
they are opposite of each other .. which means there is no net x- component .. but thats not the right answer tho..
 
  • #4
1,041
4
hi
i dont know if this has been posted or not but i am sort fo stuck on this one

What is the force F_vec on the 1 nC charge at the bottom? View Figure
Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to three significant figures.

the figure looks like this View attachment 12308

attempt to this problem:

i found out the forces individually and added them together .. but it didnt work

x-component:
6nC = 0
2nC = 5.09 x 10[tex]^{-5}[/tex] each

y-component:
6nC = 2.16 x 10[tex]^{-5}[/tex]
2nc = 5.09 x 10[tex]^{-5}[/tex] each

when i add them: 1.02 x 10 [tex]^{-5}[/tex] [tex]\hat{x}[/tex] , 7.76 x 10[tex]^{-5}[/tex] [tex]\hat{y}[/tex]

but that was wrong.. i also tried the x-comp to be 0 .. but that was wrong too

thanx in advance for the help
I see 2 things you might do differently to help.
Make sure you're doing vectors (right is +, up is +). Keep the vector notation throughout.
Check your use of scientific notation. If you're unsure, have your calculator switch it to regular decimal.
 
  • #5
I see 2 things you might do differently to help.
Make sure you're doing vectors (right is +, up is +). Keep the vector notation throughout.
Check your use of scientific notation. If you're unsure, have your calculator switch it to regular decimal.
i tried a similar question, with 1 + and 1 - 2nC charge .. and that worked out
but somehow i cant get this working ..

i am not too sure if the y - component it right as well or not .. even tho it seems right
 
  • #6
Doc Al
Mentor
44,877
1,129
they are opposite of each other .. which means there is no net x- component .. but thats not the right answer tho..
Show how you calculated the vertical components of the forces from the 2 nC charges.
 
  • #7
Doc Al
Mentor
44,877
1,129
y-component:
6nC = 2.16 x 10[tex]^{-5}[/tex]
2nc = 5.09 x 10[tex]^{-5}[/tex] each
What are the directions (signs) of these components?
 
  • #8
Show how you calculated the vertical components of the forces from the 2 nC charges.
y-comp = Esin[tex]\vartheta[/tex]
r= 5x10^-2
E = Kq[tex]_{1}[/tex]q[tex]_{2}[/tex] / r[tex]^{2}[/tex]
= (9.0x10^9)(2x10^-9)(1x10^-9)/ ( 2.5x10^-3)
E= 7.2x10^-6

y-comp = E sin(45) = 5.09x10^-6
 
  • #9
Show how you calculated the vertical components of the forces from the 2 nC charges.
What are the directions (signs) of these components?
6nC = +y, no x
2nC = +y, +x (on the right)
2nC = +y, -x (on the left)

and thats why my understand was that the x components would cancel out .. but i am not sure whats going on
 
  • #10
Doc Al
Mentor
44,877
1,129
6nC = +y, no x
2nC = +y, +x (on the right)
2nC = +y, -x (on the left)
Realize that the 2 nC charges are positive and the 6 nC charge is negative. Which way do their forces on the + 1 nC charge act?
 
  • #11
so the 2 +y .. should be -y ..?
 
  • #12
i did the calc again and i get

x = 0
y = 1.14x10^-5 ..

doe sthat seem right?
 
  • #13
Doc Al
Mentor
44,877
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so the 2 +y .. should be -y ..?
Correct. Since the charges repel, the y component of force is downward and thus negative.
 
  • #14
thank you ., thank you ... thank you..
 
  • #15
Doc Al
Mentor
44,877
1,129
y-comp = Esin[tex]\vartheta[/tex]
r= 5x10^-2
E = Kq[tex]_{1}[/tex]q[tex]_{2}[/tex] / r[tex]^{2}[/tex]
= (9.0x10^9)(2x10^-9)(1x10^-9)/ ( 2.5x10^-3)
E= 7.2x10^-6

y-comp = E sin(45) = 5.09x10^-6
What's the distance between the charges?
 
  • #16
i did the calc again and i get

x = 0
y = 1.14x10^-5 ..

doe sthat seem right?
what you said abt the charges repelling .. this answer is right ..

thank you so much once again
 
  • #17
Doc Al
Mentor
44,877
1,129
what you said abt the charges repelling .. this answer is right ..

thank you so much once again
You are most welcome.

What's the distance between the charges?
Never mind. You had written the distance squared. :smile:
 

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