Two charged spheres hitting each other

  • #1
lorenz0
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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.
 
Last edited:

Answers and Replies

  • #2
nasu
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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
lorenz0
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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
nasu
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The 1000 in front of the square root does not look right. The mass should be under the square root too.
 
  • #5
lorenz0
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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
kuruman
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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
nasu
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The sign of the term in the parenthesis is negative too. So, it's ok.
 
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Likes kuruman and String theory guy
  • #8
kuruman
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The sign of the term in the parenthesis is negative too. So, it's ok.
Oops. So it is.
 

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