# Kinetic Energy of a sphere due to uniform line of charge

• zabumafu
In summary, a small sphere with positive charge 6.00uC released from rest at a distance of 1.20cm from an infinitely long line of charge with linear charge density 4.00uC/m. The kinetic energy of the sphere when it is 4.80cm from the line of charge can be found by using the formula for change in potential, which is equal to the work done by the electric force on the sphere. By using Gauss's Law and the formula for potential difference, the kinetic energy can be calculated to be approximately -0.598J.
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.

Last edited:
Does the symbol uC stand for microCoulomb or nanoCoulomb?

TSny said:
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

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?

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.

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.

TSny said:
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

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

SammyS said:
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

## What is the formula for calculating the kinetic energy of a sphere due to a uniform line of charge?

The formula for calculating the kinetic energy of a sphere due to a uniform line of charge is KE = 1/2 * mv^2 = 1/2 * (qE)^2 * (3/5 * r), where m is the mass of the sphere, v is the velocity, q is the charge of the sphere, E is the electric field strength, and r is the radius of the sphere.

## How does the velocity of the sphere affect its kinetic energy?

The kinetic energy of a sphere due to a uniform line of charge is directly proportional to the square of its velocity. This means that as the velocity increases, the kinetic energy also increases.

## Is the kinetic energy of the sphere affected by the magnitude of the charge?

Yes, the kinetic energy of the sphere is directly proportional to the square of the charge. This means that as the charge increases, the kinetic energy also increases.

## What is the relationship between the electric field strength and the kinetic energy of the sphere?

The kinetic energy of a sphere due to a uniform line of charge is directly proportional to the square of the electric field strength. This means that as the electric field strength increases, the kinetic energy also increases.

## Can the radius of the sphere affect its kinetic energy due to a uniform line of charge?

Yes, the kinetic energy of the sphere is directly proportional to the radius of the sphere. This means that as the radius increases, the kinetic energy also increases.

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