Finding Magnitude of a Charge on Sphere

In summary, two small metallic spheres with a mass of 0.20 g each are suspended by light strings from a common point. They are given the same electric charge and come to equilibrium at an angle of 5.0 degrees with the vertical. Using Coulomb's Law and comparing the forces along the x and y axes, it is determined that the magnitude of the charge on each sphere is 7.2nC.
  • #1
cheerspens
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0

Homework Statement


Two small metallic spheres, each of mass 0.20 g, are suspended as pendulums by light strings from a common point as shown in the figure I attached. The spheres are given the same electric charge, and it is found that they come to equilibrium when each string is at an angle of 5.0 degrees with the vertical. If each string is 30.0 cm long, what is the magnitude of the charge on each sphere?


Homework Equations


Coulomb's Law


The Attempt at a Solution


I've drawn a force diagram for the first ball and I find that it's Fg=.00196, the tension, T, is .00197, and the Force of ball 2 on ball one, F21, is 1.71x10-4. When I plug into this equation: F=(ke|q|)/(r2), I get 5.1x10-9.

However, I know this is wrong because the correct answer is 7.2nC. How do I do this correctly?
 

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  • #2
Notice that F = K * q * q / r^2...
Your formula lacks one q... Maybe that's where it fell? I'm too tired to make the calculation :)
 
  • #3
Ok I've made sure that I account for both q's.
In order to solve this, do I have to break the vectors into components and set them equal to each other?
 
  • #4
Yes. It seems to me you did everything properly.
You get the tension from comparing the y-axis forces: Gravity and T*sin5.
Then by comparing the x-axis forces: T*cos5 = Kq^2 / r^2 you can isolate q.
notice that r = 0.3 * sin5 * 2.
It should work. If you still get the same answer, then the book's wrong...
 
  • #5
I'm not sure if you took the time to actually solve it (I wouldn't blame you if you didn't) but for my final answer I got 6.88nC. If the book's answer is 7.2nC would it be safe to say this is accurate?
 
  • #6
I'm getting the book's answer after trying.

You get the equation:

0.00197 * sin5 = 9 * 10^9 * q^2 / (0.3*(sin5)*2)^2)

solving it leads to q = 7.22 nano C.
 

What is the formula for finding the magnitude of a charge on a sphere?

The formula for finding the magnitude of a charge on a sphere is Q = 4πεr²V, where Q is the charge, ε is the permittivity of free space, r is the radius of the sphere, and V is the potential difference.

How do you determine the radius of a sphere when finding its charge?

The radius of a sphere can be determined by measuring the distance from the center of the sphere to any point on its surface. This distance is denoted as r in the formula Q = 4πεr²V.

What is the unit for the magnitude of a charge on a sphere?

The unit for the magnitude of a charge on a sphere is coulombs (C).

Can the magnitude of a charge on a sphere be negative?

Yes, the magnitude of a charge on a sphere can be negative. This indicates that the charge is negative and the sphere has an excess of electrons.

What factors affect the magnitude of a charge on a sphere?

The magnitude of a charge on a sphere is affected by the potential difference (V) and the permittivity of free space (ε). It is also indirectly affected by the radius (r) of the sphere, as seen in the formula Q = 4πεr²V.

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