Help with with finding the magnitude of a charge

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To find the magnitude of charge Q balancing a 1.0 kg object against another charge, the gravitational force (9.8 N) is equated to the electrostatic force using Coulomb's law. The calculations yield a charge of approximately 24.4 μC. The approach involves setting the forces equal and solving for Q, confirming that the method is correct. The final answer is validated as 24 μC. This method effectively demonstrates the relationship between gravitational and electrostatic forces.
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Homework Statement


The weight of a 1.0 kg object of charge Q is just balanced by another object of equal but opposite charge fixed to a support 74 cm above it. What is the magnitude of the charge Q (in μC)?

Homework Equations


F=ke|q1||q2| / r^2
F=ma

The Attempt at a Solution


F=(1kg)(9.8m/s^2) = 9.8

9.8=8.99e9(q^2) / .74^2 = 24.4 uC
I think that's the right way, I'm not too sure.

Thanks
 
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dasblack said:

Homework Statement


The weight of a 1.0 kg object of charge Q is just balanced by another object of equal but opposite charge fixed to a support 74 cm above it. What is the magnitude of the charge Q (in μC)?


Homework Equations


F=ke|q1||q2| / r^2
F=ma


The Attempt at a Solution


F=(1kg)(9.8m/s^2) = 9.8

9.8=8.99e9(q^2) / .74^2 = 24.4 uC

Hmmm...

Yes equating the attrative force of electromagnetism to it's weight is the right way to go about the problem

0 = (kQq / r2) - mg

mg = (kQ2 / r2)

mgr2 = (kQ2

mgr / k = Q2

sqrt{ (mgr / k) } = Q

If that is what you've done then good on you! Numerically my answer comes out at: 2.84*10-5 C

In μC , 24μC I think...

Hope this helps!
Haths
 
Haths said:
Hmmm...

Yes equating the attrative force of electromagnetism to it's weight is the right way to go about the problem

0 = (kQq / r2) - mg

mg = (kQ2 / r2)

mgr2 = (kQ2

mgr / k = Q2

sqrt{ (mgr / k) } = Q

If that is what you've done then good on you! Numerically my answer comes out at: 2.84*10-5 C

In μC , 24μC I think...

Hope this helps!
Haths

Yes, 24.4 is the answer, thank you for the help.
 
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