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Electric forces and fields

  • #1
a small 2 g ball is suspended by a 20 cm string in a uniform electric field of 1.0 x 10 to the fourth power N/C. If the ball is in equilibrium when the string makes a 15 degree angle with the vertical, what is the net charge on the ball?

I need someone to check my work for me

F=mg (mass times gravity)
F=(2)(9.8)
F = 19.6 N

E = F / q

19.6 / 1.0 x 10^4 = 1.96 x 10 ^-3

q= 1.96 x 10 ^-3

is that right at all ?
 

Answers and Replies

  • #2
nrqed
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LezardValeth said:
a small 2 g ball is suspended by a 20 cm string in a uniform electric field of 1.0 x 10 to the fourth power N/C. If the ball is in equilibrium when the string makes a 15 degree angle with the vertical, what is the net charge on the ball?

I need someone to check my work for me

F=mg (mass times gravity)
F=(2)(9.8)
F = 19.6 N

E = F / q

19.6 / 1.0 x 10^4 = 1.96 x 10 ^-3

q= 1.96 x 10 ^-3

is that right at all ?
It's incorrect because you are not taking into account the tension in the string. And you are solving as if the forces are scalars. But they are vectors. You need to draw a free body diagram, decompose all the forces into x and y components and impose that the net force along the x axis is zero and the same for the net force along y.

Before doing that, you need to specify in what direction the electric field is pointing, though.
 
  • #3
http://ourworld.cs.com/LezardV4leth/P1010062.JPG [Broken]

theres the picture in the book along with the directions

the tension on the string you mean friction?
 
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  • #4
nrqed
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LezardValeth said:
http://ourworld.cs.com/LezardV4leth/P1010062.JPG [Broken]
Ok I see now
theres the picture in the book along with the directions

the tension on the string you mean friction?
No, I mean the tension in the string holding the charge. Have you done free body diagrams? Have you done problems in which a string is attached to an object? There is a tension force in the string.


First, you have to decide the sign of the charge.

Then, you will have to draw a FBD, decompose all the forces into x and y components and then impose that [itex] \sum F_x = 0, \sum F_y=0 [/itex] and then solve for the charge.

Patrick
 
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  • #5
yes weve done FBD but I think weve only done that with collisions and kinetic energy. I dont recall doing anything with strings besides friction
 
  • #6
nrqed
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LezardValeth said:
yes weve done FBD but I think weve only done that with collisions and kinetic energy. I dont recall doing anything with strings besides friction
I don't think we are using the same language at all here:confused:

Free body diagrams show all the forces acting on an object (which may be accelerating or not, maybe in motion or at rest). There is no kinetic energy involved, and it's rare that one draws a FBD during a collision (FBD involve forces and not energy or momentum, which is maybe what you are thinking about). Also, a string has nothing to do with friction!

This problem is too hard to explain through the Internet if you haven't worked with free body diagrams a bit before, or at least decomposed forces into x and y components.
 
  • #7
lol =P I think my teacher just drew the picture and placed arrows or something. I thank you taking the time and helping me with this problem I really appreciate it

heres what I have so far http://ourworld.cs.com/LezardV4leth/13432.jpg [Broken]

can you explain to me what tension is? or is there a technical term that I can look up in my text book for that?
 
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  • #8
Hootenanny
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LezardValeth said:
can you explain to me what tension is? or is there a technical term that I can look up in my text book for that?
Tension is a force acting on a body in a direction as to produce strain (or a stretching effect).

-Hoot:smile:
 
  • #9
nrqed
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LezardValeth said:
lol =P I think my teacher just drew the picture and placed arrows or something. I thank you taking the time and helping me with this problem I really appreciate it

heres what I have so far http://ourworld.cs.com/LezardV4leth/13432.jpg [Broken]

can you explain to me what tension is? or is there a technical term that I can look up in my text book for that?
Tension is just the force exterted by the string on the charge. (here it is acting along the red line in your diagram)

Consider a simpler problem: a mass hanging from a rope (no electric force). The mass is at rest. Clearly the net force on it is zero. There is obviously the gravitational force acting downward. There must be a force acting upward to cancel this. It's obviously produced by the string which is holding the mass. The force exterted by the string on the mass is what we call the tension force. In this case, it is obviously [itex] {\vec T} = mg {\vec j} [/itex] so that [itex] \sum {\vec F} = {\vec T} + {\vec F_{grav}} = 0 [/itex]. So the magnitude of the tension force, denoted T, is equal to mg in this case.

In your example, it won't be equal to mg because there is another force. You must call the magnitude of the tension force T and treat it as an unknown.


I have to run to classes. I will write more later today.

Patrick
 
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