Charge Q is distributed uniformly over a thin rod

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SUMMARY

The discussion focuses on calculating the electric field generated by a uniformly charged thin rod with total charge Q, extending from points (a,0) to (b,0) on the x-axis. The linear charge density is defined as λ = Q / (b - a). The electric field at the origin is derived using the equation E = k * dQ / r², leading to the integral E = k * λ * ∫(dx/x²) from a to b, resulting in E = k * (Q/(b-a)) * (-1/b + 1/a) after simplification. This approach emphasizes expressing λ in terms of Q, a, and b for clarity.

PREREQUISITES
  • Understanding of linear charge density and its calculation
  • Familiarity with electric field concepts and equations
  • Knowledge of calculus, specifically integration techniques
  • Basic understanding of electrostatics and Coulomb's law
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  • Study the derivation of electric fields from continuous charge distributions
  • Learn about the application of Gauss's Law in electrostatics
  • Explore the concept of potential difference in electric fields
  • Investigate the effects of varying charge distributions on electric fields
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Charge Q is distributed uniformly over a thin rod...

Homework Statement



Charge Q is distributed uniformly over a thin rod of
lying along the x-axis between points (a,0) and (b,0)
as shown. Please answer each of the following
questions.

a) What is the linear charge density on the rod?
b) Find the electric field at the origin.


Homework Equations


linear charge density = Q / l
E = k Q / rˆ2

The Attempt at a Solution



This is what I got so far ...

dQ = lambda*dx = Qdx/(b-a)

integral of E = k dQ/rˆ2 i
= integral from a to b of k*lambda*dx/xˆ2 = k*lambda*(-1/b+1/a)
 
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Looks good. You get a nicer looking answer for b if you drop the lambda and express it in terms of the given Q, a and b. Get a common denominator for that nasty 1/a - 1/b.
 


yes, just proceed and replace lambda.
 

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