Coulomb force on a line charge

In summary, the problem involves a source charge in the form of a uniform sphere centered at the origin and a line charge with uniform charge density along the Z axis. The solution involves resolving the source charge as a point and calling the line charge L*\lambda. The vector from the source charge to the line charge is expressed as Z*Z(hat), and the integration is performed from Z0 to Z0+L with 1/Z2 as the integrand to simplify the solution.
  • #1
Kudaros
18
0

Homework Statement



We have a source charge that is a uniform sphere with a radius a (centered at origin) and uniform charge density, [tex]\rho[/tex]. There is a line charge with a length L that begins at Z0 and ends at Z0 + L (lies on the Z axis). This line charge has a uniform charge density of [tex]\lambda[/tex].

Homework Equations



Ill combine this with my attempt.

The Attempt at a Solution



First, I am resolving the source charge as a point. The sphere is of a uniform charge density and centered on the origin. So the 'q' for this source charge is [tex]\frac{4}{3}\pi[/tex]*a2*[tex]\rho[/tex].

Second, I am calling the line charge L*[tex]\lambda[/tex].

My solution so far : Fq'onq(sphere/point on line charge)= [tex]\frac{\lambda*\rho*a^{3}*L}{3*\epsilon}*\frac{\vec{R}}{R^{3}}[/tex]
My problem (assuming the above is correct) is that I am uncertain how to express the vector, R, from the source charge to the line charge.

In general, I am unsure of how to express a vector from a point to a continuous distribution of charges, or even from a continuous distribution to another.
 
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  • #2
I don't think you can treat the line charge as a point charge, as you can do with spheres.
 
  • #3
Ok I've got it.My line charge was indeed wrong.

Basically, I chose my vector to be simple. The position with respect to the source was Z*Z(hat) and I ended up integrating from Z0 to Z0+L with 1/Z2 as the integrand ( lambda is constant, was pulled out.)

Some quick simplification results in the answer in the back of the book.

I was thinking too hard about the vector I suppose.

Thank you!
 

1. What is the Coulomb force on a line charge?

The Coulomb force on a line charge is the electrostatic force exerted on a charged particle by a charged line, in accordance with Coulomb's Law. This force is directly proportional to the magnitude of the charges and inversely proportional to the distance between them.

2. How is the Coulomb force calculated on a line charge?

The Coulomb force can be calculated using the equation F = k * (q1 * q2) / r, where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, and r is the distance between them. For a line charge, the distance (r) is measured from the point on the line charge to the point where the force is being calculated.

3. What is the direction of the Coulomb force on a line charge?

The direction of the Coulomb force on a line charge depends on the relative charges of the two particles. If the two charges are of opposite sign, the force will be attractive and will act along the line connecting the two particles. If the two charges are of the same sign, the force will be repulsive and will act away from the line connecting the two particles.

4. How does the Coulomb force change as the distance between two line charges changes?

The Coulomb force decreases as the distance between two line charges increases. This is because the force is inversely proportional to the distance between the two charges. As the distance increases, the force decreases by the square of the distance.

5. How does the Coulomb force on a line charge compare to that of a point charge?

The Coulomb force on a line charge is similar to that of a point charge, except that the line charge extends in one dimension while the point charge is confined to a single point. The equation for the Coulomb force on a point charge is F = k * (q1 * q2) / r^2, where r is the distance between the two charges. For a line charge, the distance (r) is measured from the point on the line charge to the point where the force is being calculated, but the force still follows the inverse square law.

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