How Do You Calculate Tension and Charge in Electromagnetic Problems?

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SUMMARY

This discussion focuses on calculating tension and charge in electromagnetic problems involving two charges. The tension in the thread supporting Charge 1, which weighs 0.15 kg, is determined to be 1.9 N using the equation Ft = m g / cos(40). The charge on Charge 2 is calculated to be approximately 4.7 x 10-6 C, indicating that Charge 2 is positive due to the attractive force between the negative Charge 1 and the positive Charge 2.

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Homework Statement



Charge 1 (0.15 kg) in the following diagram is hanging from a light insulating thread, as shown. Charge 2 is brought close and pulls charge 1, deflecting the string away from the vertical axis. The charges have the same magnitude, but charge 1 is negative.

A) Find the tension in the thread

B) Find the charge on charge 2. Is it positive or negative? Explain your reasoning.

I'm satisfied with my answers, however, I'm almost certain I haven't "listed" everything correctly. (example, putting FT in the wrong place where a different notation should be listed...)

I'm taking notes from the tips that I've been given in previous threads (just want to make it clear that I'm actually storing this information in order to avoid having to scramble every time a new problem arises).

2igemir.jpg


Homework Equations


[/B]
Fg = m g
Fe= K q2 / r2

The Attempt at a Solution


[/B]
A)

Ft = m g
Ft = (0.15) (9.8) / cos 40
Ft = 1.9

The tension in the thread is 1.9 N

B)

Ft = k q2 / r2
Fe = (9.0 x 109) q2 / 0.42
q = √ (0.402) (1.9 sin40) / 9.0 x 109
q = 4.7 x 10-6 C

Charge 1 (given as negative) attracts Charge 2, thus charge 2 is positive.
 

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I agree: reading
Ft = m g​
on one line, and then
Ft = (0.15) (9.8) / cos 40​
on the next makes my skin crawl. The 1/cos(40) seems to come out of the blue sky. Something like
Fty = m g
| Ft | = (0.15) (9.8) / cos 40​

is more understandable. You can then continue with
Ftx = | Ft | sin(40) = k q2 / r2 etc​
 

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