Electric Field of Linear Charge along a Line Segment

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SUMMARY

The discussion focuses on calculating the electric field, E, at point P within a straight line segment of length 3L with a linear charge density λ. The relevant equation used is dE = k dQ/r², where dQ is defined as λ dL. Participants highlight the importance of correctly identifying the limits of integration and the distance r from the charge to point P. Additionally, the interaction of charges on opposite sides of point P is emphasized, as it affects the resultant electric field vector.

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  • Understanding of electric field concepts and vector analysis
  • Familiarity with calculus, specifically integration techniques
  • Knowledge of linear charge density and its implications
  • Basic principles of electrostatics and charge interactions
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  • Learn about improper integrals and techniques for evaluating them
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Students in physics, particularly those studying electromagnetism, as well as educators and anyone involved in solving problems related to electric fields and charge distributions.

phy6
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Homework Statement


Consider a straight line segment of 3L and with a linear charge density λ. Determine the electric field, E, of at point P, which is a point within the segment and along the axis. (figure attached)

Homework Equations


dE=kdQ/r^2

The Attempt at a Solution


I attempted solving it but I am stuck at the integration part because i don't know from and up to where to integrate it.
 

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I have some trouble matching your picture to the wording in the problem statement. The wording suggests that the point P lies "within" the 3L segment (distance L from the end perhaps?). The picture shows point P at the end of the 3L segment, then another segment L adjacent to that.
As far as your integration, note that your charge density lambda, is actually dQ/dL, with dL representing "delta Length". So you could say that dQ = (lambda)*dL and integrate lengths. Note that the distance r is also a length from the charge to a point. Take care to note how charges on opposite sides of the point will interact to form the resulting electric field (which is a vector).
 
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scottdave said:
your charge density lambda, is actually dQ/dL
I advise against such a notation. L is related to the overall length, a constant. An infinitesimal should represent a small change in a variable. Let x be the distance from one end of an element length dx, etc.

@phy6, if you follow the obvious path you will get two improper integrals that do not converge. Think how you might cancel out the infinities by practical considerations first.
 
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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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