Kinetic Energy of a sphere due to uniform line of charge

[zabumafu]

Homework Statement

A very small sphere with positive charge 6.00uC is released from rest at a point 1.20cm from a very long line of uniform linear charge density 4.00uC/m.What is the kinetic energy of the sphere when it is 4.80cm from the line of charge if the only force on it is the force exerted by the line of charge?

The Attempt at a Solution

Well we have not gone over this in class so I have no notes and can't seem to find a solution online. All I could think about doing was solving for the change in potential from the initial point at 1.2cm and since its at rest, all of the energy lost will be converted in kinetic.

So change in potential=-W/q with q being the charge of the particle.

Vi=initial potential
Vf=final potential

Vi=k(lamda)ln(xi)
=(8.99*10^9)(4.0*10^-6)ln(.012)
=-1.59*10^5
Vf=(8.99*10^-9)(4.0*10^-6)ln(.048)
=-1.09*10^5
dV=-4.985*10^4=-W/q

W=49.85*(6.0*10^-6)
W=K=.299J?

The answer is wrong but I have no other formulas to determine potential of a uniform line of charge.

 

[TSny]

Does the symbol uC stand for microCoulomb or nanoCoulomb?

 

[zabumafu]

Does the symbol uC stand for microCoulomb or nanoCoulomb?

microCoulomb sorry should have noticed that before and just noticed I did nano, so let me rework that however I only have 1 more attempt at the problem so I want to make sure its right. I corrected it to .299J I believe using microCoulombs

 

[TSny]

Are you sure you got the numerical factor in front of the natural log function correct? Maybe a factor of 2 or 1/2 or something?

 

[zabumafu]

I am not sure, I used the only equation involving lambda provided by our book. If your hinting that there is, I'd say by a factor of 1/2 but can't explain why other than by basing it off of other derived equations.

 

[TSny]

You should be able to consult the source from which you got your formula. Or do a web search for the potential of an infinite line of charge.

 

[zabumafu]

You should be able to consult the source from which you got your formula. Or do a web search for the potential of an infinite line of charge.

My book (the source I was using) says the potential due to a continuous line of charge is

V=(λ/4∏εo)ln[(L+(L^2+ d^2)^.5)/d] however what confused me is if L is infinitely long and ignored, that turns the natural log portion of the function into +-d/d

 

[SammyS]

My book (the source I was using) says the potential due to a continuous line of charge is

V=(λ/4∏εo)ln[(L+(L^2+ d^2)^.5)/d] however what confused me is if L is infinitely long and ignored, that turns the natural log portion of the function into +-d/d

As is the case for finite charge distributions, that potential goes to zero as d → ∞ .

However, if you let L → ∞ , then the potential is not defined for any finite value of d .

Use Gauss's Law to find the electric field a distance r, from an infinitely long line of charge with linear charge density, λ . Then use:

\displaystyle V(b)-V(a)=-\int_{a}^{b}\,\vec{E}\cdot d\vec{r}​

 

[zabumafu]

As is the case for finite charge distributions, that potential goes to zero as d → ∞ .

However, if you let L → ∞ , then the potential is not defined for any finite value of d .

Use Gauss's Law to find the electric field a distance r, from an infinitely long line of charge with linear charge density, λ . Then use:

\displaystyle V(b)-V(a)=-\int_{a}^{b}\,\vec{E}\cdot d\vec{r}​

Perfect so

dV=-λ/2πεo[ln(a)/(b)]
=[-(4.0*10^-6)/2π(8.85*10^-12)]*ln(.048/.012)
dV=-99725V
dV=U/q
U=K=-.598J which was correct