Two charged spheres hitting each other

In summary: Thanks for checking.In summary, the conversation discusses the use of conservation of energy in a problem involving gravity and electric force. The initial and final energy equations are provided, and after solving for velocity, a mistake is pointed out and corrected. The conversation also addresses a concern about a negative term under the square root in the working equation. After further examination, it is determined that the negative sign is accounted for and the solution is correct.
  • #1
lorenz0
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28
Homework Statement
Consider two small charged sphere of mass ##m=1g## and radius ##r=50\mu m##. One has charge ##q_1=6\mu C## and the other ##q_2=-6\mu C ##. They are released from rest at a distance of ##r_0=1mm##. Find the speed with which they hit each other.
Relevant Equations
##E=k\frac{q_1q_2}{r}-G\frac{m^2}{r} ##
Since the forces involved (gravity and electric force) are conservative we can use conservation of energy.
The initial energy is ##E_i= k\frac{q_1q_2}{r_0}-G\frac{m^2}{r_0} ## and the final ##E_f=mv^2+k\frac{q_1q_2}{2r}-G\frac{m^2}{2r} ## so from ##E_i=E_f ## we get ##v=\sqrt{\left(kq_1q_2\left(\frac{1}{r_0}-\frac{1}{2r}\right)+Gm^2\left( \frac{1}{2r}-\frac{1}{r_0} \right) \right)}\frac{1}{m}=1000\cdot\sqrt{ 9\cdot 10^9 \cdot(-36\cdot 10^{-12})\cdot( 10^3 -10^4 ) +6.67\cdot 10^{-11}\cdot 10^{-6}\cdot (10^4 -10^3) }m/s\approx 54,000 m/s ## but the solution to this problem says that the naswer should be ##\approx 1700 m/s ## and I don't see what I am doing wrong so I would appreciate some feedback on my solution, thanks.
 
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  • #2
You can drop the gravitational effect. Maybe so you can figure out the mistake. As it is written (with or without gravity), you get a negative number under the square root. How did you get your answer? You need to show what you have done if you expect someone to see what is wrong
 
  • #3
nasu said:
You can drop the gravitational effect. Maybe so you can figure out the mistake. As it is written (with or without gravity), you get a negative number under the square root. How did you get your answer? You need to show what you have done if you expect someone to see what is wrong
Thanks for your interest in my question. I had already included my work, the only thing left to do was to show the formula I had derived with the numbers plugged in, which I have now done. The number under the square root is actually positive, according to my calculator.
 
  • #4
The 1000 in front of the square root does not look right. The mass should be under the square root too.
 
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  • #5
nasu said:
The 1000 in front of the square root does not look right. The mass should be under the square root too.
ah, of course! I don't know how I didn't see that before, I was thinking I hadn't understood the situation correctly but it was just a typo, many thanks.
 
  • #6
You also messed up the energy conservation equation. Are you not worried about the negative sign under the radical in the factor ##(-36\times 10^{-12})##? Please check your work again.
 
  • #7
The sign of the term in the parenthesis is negative too. So, it's ok.
 
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  • #8
nasu said:
The sign of the term in the parenthesis is negative too. So, it's ok.
Oops. So it is.
 

1. What happens when two charged spheres hit each other?

When two charged spheres hit each other, they will experience a force of attraction or repulsion depending on the sign of their charges. If the charges are of the same sign, they will repel each other, and if they are of opposite signs, they will attract each other.

2. How does the speed of the spheres affect the collision?

The speed of the spheres will affect the force of the collision. If the spheres are moving at a high speed, the force of the collision will be greater, resulting in a larger change in the direction of their movement. However, if the spheres are moving at a low speed, the force of the collision will be smaller, resulting in a smaller change in their direction of movement.

3. Can the charges of the spheres change during the collision?

Yes, the charges of the spheres can change during the collision. If the spheres are made of conductive materials, the charges can redistribute between them, resulting in a change in their overall charge. This is known as the triboelectric effect.

4. What factors affect the outcome of the collision?

The outcome of the collision between two charged spheres can be affected by several factors, including the magnitude and sign of the charges, the speed and direction of the spheres, and the distance between them. The mass and size of the spheres can also play a role in the collision.

5. Is the collision between two charged spheres perfectly elastic?

No, the collision between two charged spheres is not perfectly elastic. Some energy will be lost during the collision due to factors such as friction and heat. However, the collision can be considered mostly elastic if the spheres are made of non-conductive materials and the collision occurs in a vacuum.

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