Discover the Charge on Two Suspended Charged Balls with E and M Help!

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SUMMARY

The discussion centers on calculating the charge on two suspended charged balls using electrostatics and trigonometry. The problem involves two balls, each with a mass of 30g, suspended by 10cm strings, forming a 60-degree angle at equilibrium. The correct approach includes using Coulomb's law (F = k * q^2 / r^2) and balancing the gravitational force with the electrical force, taking into account the geometry of the setup. The initial calculation of 1.33x10^-6 C requires verification through proper application of trigonometric principles and force balance.

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  • Understanding of Coulomb's Law (F = k * q^2 / r^2)
  • Basic trigonometry for calculating angles and distances
  • Knowledge of gravitational force equations
  • Concept of equilibrium in physics
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Students studying physics, particularly those focusing on electrostatics and mechanics, as well as educators looking for practical examples of charge calculations in equilibrium scenarios.

angiecakes
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E and M help!

I got 1.33x10^-6 for this question, can someone please verify/check it?

"two charged balls are suspended on light 10cm strings. Each has a mass of 30g. In equilibrium, the total angle between the strings is 60 degrees. Assuming the balls are equally charged, what is the charge on each ball?"

:(

Thanksss!
 
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I'm assuming the balls would be right next to each other if hanging straight down?

Did you
A)Use trigonometry to figure out the distance between the balls at equilibrium?
B)Use F=c*q^2/r^2 to figure out the force on the balls in terms of q?
C)Use trig to figure out the force from gravity acting in the "swinging" direction on the balls?
D)Use trig to figure out the electrical force(in terms of q still)acting against the "swinging" direction?
E)Set those two equal and solve for q?

What you probably did wrong, if you did something wrong(I didn't check the math)is do F=c*q^2/r^2 just like that, which is the force going in a straight line between the two, and set that equal to the gravitational force that's NOT going in a straight line between the two, and it's not going to work
 
I think it would help us if you posted exactly what you did to get the answer you've stated angiecakes.
 

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