Electric forces/static equilibrium doozy

[etc]

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?

 

[Andrew Mason]

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:

$$F_{horiz} = T_{horiz} = \frac{kQ^2}{d^2}$$ (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:
$$T_{vert} = m_{ball}g$$ (2)

Since:

$$T_{horiz} = T_{vert} / tan\theta$$ (3)

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

$$Q^2 = T_{horiz}d^2/k$$

I get
$$T_{vert} = 1.96 \times 10^{-2}N.$$

$$T_{horiz} = 3.4 \times 10^{-2}N.$$

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

$$Q = \sqrt{1.02 \times 10^{-12}} C. = 1.01 \times 10^{-6} C.$$

AM

 

[etc]

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.

 

[Andrew Mason]

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

 

[hhegab]

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