Charges are connected and then kicked by chuck norris

[1MileCrash]

Homework Statement

Two identical conducting spheres, fixed in place, attract each other with a force of 0.111 N when their center to center separation is 45.00 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres have a net positive charge and repel each other with an electrostatic force of 0.043 N. What was the initial negative charge on one of the spheres, and what was the initial positive charge on the other?

Homework Equations

Coulomb's Law

The Attempt at a Solution

I posted a problem nearly EXACTLY like this and I still don't understand.

First, since they are attracting initially I know they have opposite signs. To be mathematically consistent, I will then call this force in the negative direction.

$-.111 = kq_{1}q_{2}r^{-2}$

I also know that after being connected, both spheres have a charge of

$q_{f} = 0.5(q_{1} + q_{2})$

So then I can use Coulomb's Law with the final force,

$0.43 = kq^{2}_{f}r^{-2}$

Which allows me to solve for that final charge.

$9.84x10^{-8} = q_{f}$

Then I can use this numerical value to solve for one of the initial forces.

$1.96x10^{-7} - q_{1} = q_{2}$

Then I can substitute this into the initial force equation for a quadratic.

$0 = 1.96x10^{-7}q_{1} - q^{2}_{1} + 2.5x10^{-14}$

Then I can use the quadratic formula to get the two solutions,

-8.8x10^-8, and 2.84x10^-7.

Then I can hit submit to waste one of my attempts because I am doing it wrong.

 

[Harrisonized]

repel each other with an electrostatic force of 0.043 N.

So then I can use Coulomb's Law with the final force,

$0.43 = kq^{2}_{f}r^{-2}$

Because of this?

 

[1MileCrash]

No, typo.

 

[Harrisonized]

Well, here's how I would have done it.

F_i = kq₁q₂r^-2

F_f = kq_f²r^-2
↓ q_f = 0.5(q₁+q₂)
= 0.25k(q₁+q₂)²r^-2

q₂ = (4F_f k^-1r²)^1/2 -q₁

F_f = 0.25k(q₁²+2q₁q₂+q₂²)r^-2
= 0.25k(q₁²+q₂²)r^-2+0.5F_i

4(F_f -0.5F_i)k^-1r² = q₁²+q₂²
4(F_f -0.5F_i)k^-1r² = q₁²+((4F_f k^-1r²)^1/2 -q₁)²
4(F_f -0.5F_i)k^-1r² = q₁²+(q₁²-2(4F_f k^-1r²)^1/2q₁+4F_f k^-1r²)
2q₁²-2(4F_f k^-1r²)^1/2q₁+4F_f k^-1r²-4(F_f -0.5F_i)k^-1r² = 0
q₁²-(4F_f k^-1r²)^1/2q₁+F_i k^-1r² = 0

q₁ = (4F_f k^-1r²)^1/2 ± √(4F_f k^-1r² -4F_i k^-1r²)/2
q₁ = (4F_f k^-1r²)^1/2 ± ((F_f -F_i)k^-1r²)^1/2
q₁ = ((4F_f ±(F_f -F_i))k^-1r²)^1/2

Then plug in numbers. The reason it is better this way is that now you have solved this problem for all possible given numbers, rather than only the one specifically given. It's also easy to check if the equation makes dimensional sense (if it fits the form of the Coulomb's law).

For this problem:
F_f = 0.043N and F_i = -0.111N. Since k is normally given in Nm²C^-2, if you convert the r = 45.00 cm to meters, then the units of q is in Coulombs.

You should go back and check your previous "problem nearly EXACTLY like this" using this solution.

 

[Nickie]

It seems like you are on the right track with your approach to solving this problem. However, there are a few mistakes in your calculations. First, the initial force equation should be negative since the spheres are initially attracting each other. Also, when using Coulomb's Law, it is important to use the correct distance between the spheres, which in this case is the center-to-center separation of 45.00 cm. Additionally, when solving for the final charge, you should use the net force of 0.043 N, not 0.43 N. Finally, when solving for the initial charge on one of the spheres, you should only use the positive solution from the quadratic formula, as the negative solution would result in a negative charge and we know that both charges must be positive. Keep working on your approach and make sure to double check your calculations to find the correct solution.