Calculating the electric field due to a wire of finite length

In summary, the conversation discusses determining the components of the electric field near one end of a uniformly charged wire, using integration and the equation E=KQ/R^2. The y and x components of the electric field are calculated using sin(t) and cos(t), respectively, and integration is used to solve for the components.
  • #1
aftershock
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Homework Statement



Suppose a uniformly charged wire starts at point 0 and rises vertically along the positive y-axis to a length L. Determine the components of the electric field Ex and Ey at point (x,0). That is, calculate [tex]\vec{}E[/tex] near one end of a long wire, in the plane perpendicular to the wire.

Homework Equations



E= KQ/R^2 with a point charge. So I'll be using integration as well.


The Attempt at a Solution



Ditching the vector for now and focusing on magnitude I have

dE = K (dQ/R^2)

[tex]\lambda[/tex]dL = dQ (where lambda is the linear charge density)

dE = k[tex]\lambda[/tex]dL/(x^2 +y^2)

pretty much can't figure out how to complete the integration...
 
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  • #2
dL should be dy

you need to break the E field into components

the y component of the E field is (k*dq)/(x^2+y^2) * sin(t)
the x component of the E filed is (k*dq)/(x^2+y^2) * cos(t)

once you get there, use a U substitution and the integral should be trivial
 

1. What is the equation for calculating the electric field due to a wire of finite length?

The equation for calculating the electric field due to a wire of finite length is:
E = kλ/2πr, where E is the electric field strength, k is the Coulomb's constant, λ is the linear charge density of the wire, and r is the distance from the wire.

2. How does the electric field vary along the length of the wire?

The electric field varies inversely with the distance from the wire. This means that the closer you are to the wire, the stronger the electric field will be, and the farther you are from the wire, the weaker the electric field will be.

3. What is the direction of the electric field due to a wire of finite length?

The direction of the electric field will be perpendicular to the wire at all points. This means that the electric field lines will point away from the wire if the wire is positively charged, and towards the wire if the wire is negatively charged.

4. Can the electric field due to a wire of finite length be negative?

Yes, the electric field can be negative. This will occur if the wire has a negative charge, as the direction of the electric field will be towards the wire. However, the magnitude of the electric field will still decrease with distance from the wire.

5. How does the electric field due to a wire of finite length compare to that of an infinite wire?

The electric field due to a wire of finite length will be stronger near the ends of the wire, as compared to an infinite wire where the electric field is constant along its length. Additionally, the electric field due to a finite wire will decrease more rapidly with distance as compared to an infinite wire.

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