Help with Coulomb's Law: Solving for Charges on Identical Spheres

In summary, in attempt 1 of Coulomb's law, the negative charge of one of the spheres is not equal to the positive charge of the other sphere. In attempt 2, the positive charge of one of the spheres is not equal to the negative charge of the other sphere. However, using the original equation 1 and method 2 of solving for q1 and q2, the equation correctly gives the correct result of q_1 q_2=3.00(10^{-12})
  • #1
Saladsamurai
3,020
7
!Coulomb's Law!

So I have already solved this one before, but I was redoing it fir practice when I encountered something that is troubling me. Depending on how I choose to solve my two equations, I get different results. Surely I am doing something wrong, but I cannot see it.

Problem
Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of .108 N when their center-to-center separation is 50 cm. The spheres are then connected by a thin conducting wire. When the wire is disconnected, the spheres repel each other with an electrostatic force of .036 N. Of the initial charges on the spheres, with a positive net charge, what was the (a) negative charge of one of them and (b) the positive charge of the other?

Now I have used conservation of charge for after they connect and I end up with two equations and two unknowns:

[tex]F_e=\frac{kq_1q_2}{r^2}\Rightarrow q_1q_2=3.00(10^{-12})[/tex] (1)

[tex]F_e'=k\frac{(\frac{q_1+q_2}{2})^2}{r^2}\Rightarrow q_1+q_2=2.00(10^{-6})[/tex] (2)

Attempt 1:

If I solve (2) for q_1 then [itex]q_1=2(10^{-6})-q_2[/itex]

plugging the above into (1) [itex] -q_2^2+2(10^{-6})q_2-3(10^{-12})=0[/itex] gets me a nonreal answer.

Attempt 2:

BUT if I solve (1) for q_1 then [itex]q_1=\frac{3(10^{-12})}{q_2}[/itex] and plugging that into (2) I get [itex]q_2^2-2(10^{-6})q_2-3(10^{-12})=0[/itex] which solves correctly.

I am consistently of by a sign in the first attempt. Can anyone see what the problem is?
 
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  • #2
Saladsamurai said:
Attempt 2:

BUT if I solve (1) for q_1 then [itex]q_1=\frac{3(10^{-12})}{q_2}[/itex] and plugging that into (2) I get [itex]q_2^2-2(10^{-6})q_2-3(10^{-12})=0[/itex] which solves correctly.

Wouldn't that be + (plus) 3e-12? (Resulting in an equivalent expression to than in attempt 1.)
 
  • #3
neutrino said:
Wouldn't that be + (plus) 3e-12? (Resulting in an equivalent expression to than in attempt 1.)

No. Attempt 2 is correct. Attempt 1 is not.
 
  • #4
Awaking this thread. I just went back to try this again and have the same problem... so conceptually, something is off.
 
  • #5
Hi Saladsamurai,

In your original post you have an error in equation 1. There is then another mathematical error in method 2 that counteracts the error in equation 1 and gives the right answer.

For your equation 1, you have:

[tex]F_e=\frac{kq_1q_2}{r^2}\Rightarrow q_1q_2=3.00(10^{-12})[/tex]

but that is not right. The formula for Coulomb's law is normally given in terms of magnitudes:

[tex]F_e=\frac{k |q_1||q_2|}{r^2}[/tex]

and so you get:

[tex]
|q_1 q_2|=3 \times 10^{-12}
[/tex]

You know that either q1 or q2 is negative, and the other is positive, so your equation 1 should be:

[tex]
q_1 q_2 = - 3 \times 10^{-12}
[/tex]


In your method 2, the quadratic equation you get is the correct equation to get; however, it is not what you get from your original equations 1 and 2 in your post. If you use your original equations, the [itex] 3\times 10^{-12}[/itex] term turns out to be positive, which gives the same (nonreal) answer as method 1.
 

1. What is Coulomb's Law?

Coulomb's Law is a fundamental law in physics that describes the electrostatic interaction between two charged particles. It states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

2. How is Coulomb's Law used in science?

Coulomb's Law is used in various fields of science, such as physics, chemistry, and engineering, to calculate the force between two charged particles. It is also used to understand and analyze the behavior of electrically charged objects and to design electrical circuits and devices.

3. What are the units of measurement for Coulomb's Law?

The units of measurement for Coulomb's Law depend on the system of units being used. In the SI system, the units are newtons (N) for force, coulombs (C) for charge, and meters (m) for distance. In the CGS system, the units are dynes (dyn) for force, electrostatic units (esu) for charge, and centimeters (cm) for distance.

4. What is the difference between Coulomb's Law and Newton's Law of Universal Gravitation?

While both laws describe the force between two objects, Coulomb's Law applies to charged particles, whereas Newton's Law of Universal Gravitation applies to all objects with mass. Additionally, Coulomb's Law is based on the inverse square relationship, while Newton's Law of Universal Gravitation is based on the inverse square law.

5. Can Coulomb's Law be applied to objects with non-point charges?

Coulomb's Law can be applied to objects with non-point charges by considering the charge distribution of the object. The total force between two objects is the sum of the forces between each individual charge on the objects. This can be calculated by integrating Coulomb's Law over the charge distribution.

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