Kinetic and potential energy of a particle attracted by charged sphere

In summary, the conversation discusses a problem involving a particle at point A with charge and an unmovable sphere of radius at point B with a volumic charge density . The distance between the two points is and both objects have opposite charges. The goal is to find the speed of the particle when it reaches the sphere. To do this, the potential energy is equated to the kinetic energy and solved for the speed. However, the solution differs from the expected result and it is discovered that the potential energy lost is not but rather due to the opposite charges. Once this is
  • #1
fatpotato
Homework Statement
Find the speed of a particle initially at rest when put next to a charged sphere.
Relevant Equations
Potential energy
Kinetic energy
Hello,

I have a particle at point A with charge , and an unmovable sphere of radius at point B with a volumic charge density . The distance from particle A to the centre of the sphere in B is . Both objects have opposed charges, so, the particle in A, initially at rest, is attracted to the charged sphere and reaches it with a certain speed with modulus , which I have to find.

First, I suppose the charge of the sphere is simply .

Then, I suppose that the electric potential energy will be entirely converted into kinetic energy , when the particle reaches the sphere. I assume (and it might be wrong) that the sphere can be considered as a particle, so I consider that the two objects will touch after the particle travels a distance although in reality it touches the sphere after traveling a distance .

In the end, it boils down to So I solve for : while removing my negative sign due to the product to avoid a complex result.

However, my answer differs from the correction. Speed should be , but I get using the following numerical values :
, , , , ,

Can anyone pinpoint what I am doing wrong please?

Thank you
 
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  • #2
fatpotato said:
In the end, it boils down to So I solve for : while removing my negative sign due to the product to avoid a complex result.
This is not right. When the particle collides with the sphere, it is still a distance from the centre of the sphere, hence still has some potential energy.

And, in fact, the potential energy is zero at infinity, which is why you had an incorrect negative sign to get rid of!
 
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  • #3
Further to @PeroK's comment, the potential energy lost is not . That is the PE required to be added to move the particle away to infinity. If we define the PE at infinity as zero (which is usual) then the PE at distance r is .
If a particle moves from distance r away to distance s away the lost PE is .
 
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  • #4
Of course!

By taking into account that the particle still has potential energy of when touching the sphere I get to the correct result.

Thank you PeroK!

Edit : just saw the answer from haruspex : thank you!
 
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  • #5
haruspex said:
Further to @PeroK's comment, the potential energy lost is not . That is the PE required to be added to move the particle away to infinity. If we define the PE at infinity as zero (which is usual) then the PE at distance r is .
If a particle moves from distance r away to distance s away the lost PE is .
As opposite charges attract, the negative sign is not needed for electrostatic potential.
 
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  • #6
haruspex said:
Further to @PeroK's comment, the potential energy lost is not . That is the PE required to be added to move the particle away to infinity. If we define the PE at infinity as zero (which is usual) then the PE at distance r is .
If a particle moves from distance r away to distance s away the lost PE is .
The potential energy is negative because charges have opposite signs. You don't have to put the minus sign there. If the charges had same sign, the test charge would move by the electric force to infinity, no need to add energy to move it to infinity.
 
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  • #7
PeroK said:
As opposite charges attract, the negative sign is not needed for electrostatic potential.
Yes, I was thinking in terms of gravitational potential energy and failed to consider the charge signs. But the principle that the PE change takes the form 1/s-1/r stands.
 
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