Finding the charge given distance apart + some more

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In summary, two spheres with masses of 50.0 mg each are suspended 50.0 cm apart from a common point by massless threads. One sphere has a charge twice that of the other, causing the strings to make an angle of 6 degrees. Using Newton's principles and the forces acting on each sphere, it can be determined that the lesser charged sphere has a charge of 6.25 x 10^-8 Coulombs.
  • #1
thursdaytbs
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Two spheres, each having a mass of 50.0 mg, are suspended from a common point by massless threads 50.0 cm long. One of the spheres has been given a charge twice that of the other. The strings make an angle of 6 degrees.

What are the charges on each of the spheres?

What kind of charge is on each sphere (Positive or Negative) ? Explain.

First, I said that the distance between the two spheres is 0.0523m since you can form a right triangle using 0.5m [50cm] as the hypotonose, and 3 degrees as an angle. and multiplying that answer by 2 giving me the distance between the two spheres.

From that I know one sphere = 2q, while the other sphere = q.

So... F = (K)(q)(2q) / (r^2)

From there I'm stuck, since you have two unknowns (charge 'q' and force 'F')


I'm thinking of using F=ma = 0 (a=0), then finding using a Free Body diagram to find the force on each sphere and adding them and setting them to zero?

So sphere 1 and 2's forces = mg, tension, Electric Force (F/q)
where the Forces are equal?

I'm not too sure, can anyone give me any suggestions? Thanks for any help.
 
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  • #2
1.The 2 charges have to have opposite charge signs,else they wouldn't repel each other and would not make an angle of 6°.
2.The charges need to be in equilibrium,which means they have total acceleration zero,this meaning that the net force acting one each other is zero.
3.Write Newton's principle for each charge,knowing the 3 forces that act on each of them.
4.Use projections of forces on axes of coordinates conveninetly chosen and geomtry of the figure to find the unknowns.


Daniel.
 
  • #3
So finding exactly whether it's both positive or negative can not be done?

Also, would my equations look like this?

er... would the vectors added be: m1g, m2g, k(2q)(q) / r^2, k(2q)(q) / r^2, T1, and T2?

add them as vectors and set them equal to zero and solve for q?
 
Last edited:
  • #4
I got it. Thank you.

I did F = ma = 0.

and one sphere = equilibrium, created a right triangle with mg, Tension, and Electrical force. Knowing [mg] and the angles, I solved for the Electrical Force.

Then set that = to (k)(2q)(q) / r^2, where r = 0.0523 and k = 9x10^9.

q = 1.96 x 10^-9
2q = 3.93 x 10^-9
 
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  • #5
thursdaytbs said:
I got q = 6.25 x 10^-8C
Me too. This is the charge on the lesser charged sphere.
 
  • #6
Galileo said:
Me too. This is the charge on the lesser charged sphere.

The problem with that was that I had [mg] wrong. For the mass, I used the gram value of the mass instead of the kilogram value.
 

What is the formula for finding the charge given distance apart?

The formula for finding the charge given distance apart is F = (k * q1 * q2) / d^2, where F is the force between two charges, k is the Coulomb's constant, q1 and q2 are the magnitudes of the two charges, and d is the distance between them.

How does distance affect the magnitude of the force between two charges?

The magnitude of the force between two charges is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases, and vice versa.

What is the unit of charge in this equation?

The unit of charge in this equation is Coulomb (C). It is a derived unit in the International System of Units (SI) and is defined as the amount of electric charge carried by a current of 1 ampere in 1 second.

Can the formula be used for charges of any magnitude?

Yes, the formula for finding the charge given distance apart can be used for charges of any magnitude, whether they are positive or negative. However, the direction of the force will depend on the sign of the charges.

Can this formula be applied to non-point charges?

Yes, the formula can be applied to non-point charges as long as the distance between them is measured from their centers of mass. In this case, the charges can be treated as point charges for the purpose of calculating the force between them.

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