Field Due to Continuous Distribution of Charge

In summary: Integrating from x = (l1 + D) to x = 0 and finding the result.In summary, Coulomb force between line charges is expressed as:\frac{\lambda_{1}}{4\pi\epsilon}\int\frac{dq}{r^2}Left limits: x = DRight limits: x = (l1+D)In the expression of E at the starting point of l2, the distance l1 must appear which is missing in your expression.Try this one.dE = k*lambda1*dx/(D+x)^2.Integrate it from x = 0 to x = l1 and find E.Wait, is x the distance from
  • #1
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Homework Statement



Coulomb force between line charges: a rod of length l1 with line charge density λ1 and a rod of length l2 with line charge density λ2 lie on the x axis. Their ends are separated by a distance D as shown in the figure.

(a) What is the force F between these charges?

diagram: http://ocw.mit.edu/NR/rdonlyres/Physics/8-022Fall-2004/3A772032-6B74-4D2D-A550-8F0ECFECEDBC/0/pset1.pdf

#7


Homework Equations



E = [tex]\frac{1}{4\pi\epsilon}[/tex][tex]\int\frac{dq}{r^2}[/tex]
F = [tex]\int E dq[/tex]



The Attempt at a Solution


So, first I decided to find the field at a point a distance D from the end of line 1. Using the standard x coordinate system, I placed line 1 such that its endpoints are 0, [tex]l_{1}[/tex].

E = [tex]\frac{1}{4\pi\epsilon}[/tex][tex]\int\frac{dq}{r^2}[/tex]

Limits of integration being (0,

Using this and dq = [tex]dl_{1}[/tex][tex]\lambda_{1}[/tex], all I need to do is find a function for r in terms of l, which is the distance from 0. Which would be ([tex]l_{1}[/tex] + D) - l.

I renamed ([tex]l_{1}[/tex] + D) as the variable a to make the integration simpler. So now I have:

E = [tex]\frac{\lambda_{1}}{4\pi\epsilon}[/tex][tex]\int \frac{dl}{(a - l)^2}[/tex]

which is just [tex]\frac{\lambda_{1}}{4\pi\epsilon} * [/tex][tex]\frac{1}{a-l_{1}}[/tex]

and because a = d + [tex]l_{1}[/tex]

I get the E Field being E = [tex]\frac{\lambda_{1}}{4d\pi\epsilon}[/tex]

Is this correct so far? Clearly my success on the second part depends on that because all I have to do is just integrate the field over the infinitesimal segments of charge over the second line's length yes? And to find that distance d as a function of l it's just ( l - length 1), where l is the distance from the 0 point. I'm just kind of shaky on the first part, finding the field, that's all.
 
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  • #2
In the expression of E at the starting point of l2, the distance l1 must appear which is missing in your expression.
Try this one.
dE = k*lambda1*dx/(D+x)^2.
Integrate it from x = 0 to x = l1 and find E.
 
  • #3
Wait, is x the distance from l1? Because then D + x is the distance from some point on the line to P2 which is what i want. Because my expression actually has d in it? It's just "a" is in terms of D, and i defined my distance variable as from x = 0. I'm not sure, it looks like x+d only works if x is the distance from l1. Or am I misunderstanding something here?
 
  • #4
You want to find the field due to rod1 at the starting point of rod 2.
So E = k*lambda1*[-(1/(x+D)]. Find the value of E taking the limits from x = D to x= (l1 +D)
 

Related to Field Due to Continuous Distribution of Charge

1. What is the concept of "Field Due to Continuous Distribution of Charge"?

The concept of "Field Due to Continuous Distribution of Charge" refers to the electric field that is created by a continuous distribution of charge, such as a charged wire or a charged plate. This field is a vector quantity that describes the force that a test charge would experience at any given point in space due to the presence of the continuous distribution of charge.

2. How is the electric field calculated for a continuous distribution of charge?

The electric field for a continuous distribution of charge is calculated using Coulomb's law, which states that the electric field at a given point is equal to the product of the charge density at that point and the distance from the point to the charge squared. This calculation can be done using calculus and the formula for the electric field due to a point charge.

3. What factors affect the strength of the electric field due to a continuous distribution of charge?

The strength of the electric field due to a continuous distribution of charge is affected by the magnitude and distribution of the charge, as well as the distance from the charge. The electric field strength decreases with increasing distance from the charge and increases with increasing charge density.

4. How does the electric field due to a continuous distribution of charge compare to that of a point charge?

The electric field due to a continuous distribution of charge is similar to that of a point charge in that it follows the same inverse square law. However, the electric field from a continuous distribution of charge can vary in magnitude and direction at different points, while a point charge produces a constant electric field in all directions.

5. What are some real-life examples of continuous distributions of charge?

Some real-life examples of continuous distributions of charge include power lines, which have a continuous distribution of charge along their length, and the surface of a charged conductor, such as a metal sphere. Additionally, the Earth's magnetic field can be considered a continuous distribution of charge, as it is created by the movement of charged particles within the Earth's core.

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