Electric forces/static equilibrium doozy

Peace!In summary, two identical small spheres of mass 2.0 g are suspended by a light, flexible fishing line attached to a ceiling hook, with identical electric charges on each sphere. In static equilibrium, with an angle of 30 degrees between the string halves and the "horizon," the magnitude of the charges on each sphere can be calculated using the horizontal component of the tension in the line and the coulomb force equation. After several attempts, the calculated answer is 1.01x10^-6 C, but the expected answer is 1.2x10^-7 C. Further verification and double-checking of the problem and inputs is needed.
  • #1
etc
27
0
question:

two identical small spheres of mass 2.0 g are fastened to the ends of a 0.60 m long light, flexible, insultation fishling. The fishline is suspended by a hook in the ceiling at its exact centre. The spheres are each given an identifical electric charge. They are in static equilibrium, whith an angle of 30 degrees between the string halves and the "horizon." Calculate the magnitude of the charges of each spehere.


what I've done:
i calculated the y force necessary for each charge and i got 0.196 N. therefore i got the x charges to be 0.339 N. the distance between the two speheres is trig'd 0.5196 m. because it's in static equilibrium i figured the x charges must equal the force between the two speheres. after subbing my force into the "f=k2(q)/r^2" equation i got q to equal 5.085x10^12.

but i don't think I've got it right. can someone verify please?
 
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  • #2
etc said:
question:

two identical small spheres of mass 2.0 g are fastened to the ends of a 0.60 m long light, flexible, insultation fishling. The fishline is suspended by a hook in the ceiling at its exact centre. The spheres are each given an identifical electric charge. They are in static equilibrium, whith an angle of 30 degrees between the string halves and the "horizon." Calculate the magnitude of the charges of each spehere.

The force between the two spheres is the horizontal component of the tension in the line and is provided by the coulomb force:

[tex]F_{horiz} = T_{horiz} = \frac{kQ^2}{d^2}[/tex] (1)

where d is the separation. The separation is the distance between the centers of the spheres but because they are small you can ignore their radius and have d = .60 cos(30) = .52 m.

We can measure that force because the vertical component of the tension in the fishingline supports the weight of each ball which is:
[tex]T_{vert} = m_{ball}g[/tex] (2)

Since:

[tex]T_{horiz} = T_{vert} / tan\theta[/tex] (3)

just substitute Tvert from (2) into (3). Then work out Q by substituting Thoriz = into (1).

[tex]Q^2 = T_{horiz}d^2/k[/tex]

I get
[tex]T_{vert} = 1.96 \times 10^{-2}N.[/tex]

[tex]T_{horiz} = 3.4 \times 10^{-2}N.[/tex]

[tex]Q^2 = 3.4 \times 10^{-2} \times (.52)^2/9 \times 10^9[/tex]

[tex]Q = \sqrt{1.02 \times 10^{-12}} C. = 1.01 \times 10^{-6} C.[/tex]

AM
 
Last edited:
  • #3
should the final answer be to the -06 instead of -04?

if so, that's the answer i got after trying the problem another few times. but the answer that we're supposed to haveis 1.2x10^-7 C ... which really has me baffled.
 
  • #4
etc said:
should the final answer be to the -06 instead of -04?
Yes. I have edited the answer above.

if so, that's the answer i got after trying the problem another few times. but the answer that we're supposed to haveis 1.2x10^-7 C ... which really has me baffled.
It has me baffled too. I think the formula is correct. Just plug in the numbers. 2 grams is .002 Kg. Units in MKS. Maybe there is a mistake in the arithmetic, but i don't see it.

AM
 
  • #5
Peace!

I did the problem twice and I had the same answer as that of A M.
Either you gave us a wrong input somewhere, or that the answer you have is not accurate...Please check the problem and the inputs in it.
I have aslo tried to solve it assuming that a charge Q was divided equally between the two charges (that is Q/2 on each ball) and did not have the answer you talk about.

hhegab
 

What is an electric force?

An electric force is a type of force that is caused by the attraction or repulsion of electrically charged particles. It is one of the fundamental forces of nature and plays a crucial role in many everyday phenomena, such as electricity and magnetism.

How do electric forces affect static equilibrium?

Electric forces can affect static equilibrium by exerting a force on an object, causing it to either move or remain stationary. If the forces are balanced, the object will remain in static equilibrium. However, if the forces are unbalanced, the object will experience a net force and will accelerate in the direction of the larger force.

What is a doozy in the context of electric forces/static equilibrium?

In the context of electric forces/static equilibrium, a doozy refers to a particularly difficult or challenging problem or situation. It could involve complex calculations or require a deep understanding of the principles of electric forces and static equilibrium to solve.

What factors can affect the strength of electric forces?

The strength of electric forces is affected by several factors, including the distance between the charged particles, the magnitude of the charges, and the medium between the particles. The presence of other charged particles or objects in the vicinity can also influence the strength of electric forces.

How is static equilibrium achieved in an electrically charged system?

Static equilibrium is achieved in an electrically charged system when the net force on all the charged particles is zero. This can be achieved by balancing the forces of attraction and repulsion between the particles or by grounding the system, which allows for the charges to distribute evenly and cancel out any net force.

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