Calculating potential from a nonuniform linear charge distribution.

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SUMMARY

The discussion focuses on calculating the electric potential from a nonuniform linear charge distribution on a thin plastic rod with a charge density defined as λ = cx, where c = 28.9 pC/m², length L = 12.0 cm, and distance d = 3.00 cm from one end. The user initially attempted to integrate the potential using the formula V = ∫(kcx)/(x+d) dx but arrived at an incorrect value of 156 Volts, while the correct answer is 18.6 mV as per the textbook. The user later confirmed the correct potential using a TI-84 Plus calculator, yielding a result of 0.01863 Volts.

PREREQUISITES
  • Understanding of electric potential and charge distributions.
  • Familiarity with calculus, specifically integration techniques.
  • Knowledge of the Coulomb constant (k) and its application in electrostatics.
  • Experience with computer algebra systems, such as Microsoft Mathematics.
NEXT STEPS
  • Review the integration of electric potential for nonuniform charge distributions.
  • Study the application of the Coulomb constant in electrostatic calculations.
  • Learn how to use TI-84 Plus for solving integrals related to electric potential.
  • Explore the differences between linear charge density units (C/m vs. C/m²).
USEFUL FOR

Students in physics or electrical engineering, particularly those studying electrostatics and electric potential calculations, as well as educators seeking to clarify concepts related to charge distributions.

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Homework Statement


The thin plastic rod has length L, and a nonuniform linear charge density λ = cx. With V = 0 at infinity, find the electric potential (in V) at point P1 on the axis, at distance d from one end.
c = 28.9 pC/m^2
L = 12.0cm
d = 3.00 cm

Now, from what I can tell the left side of the rod is placed at the origin, and p1 is a distance of d from the left end.

Homework Equations



V = (kq)/r (potential of the point charge).

[itex]V = \int \frac{kcx}{x+d} dx[/itex].

k is the Coulomb constant.

The Attempt at a Solution



Now I assumed using the above equation to integrate the potential over 0 to L would give the solution. However, when I checked the back of the book I was mistaken. Is it possible I missed something? I am treating cx dx as an point charge and summing over the potential each one produces at the point in question.

I get using the equation 156Volts while the book claims the answer to be 18.6mV.
 
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Why is c C/m^2? It should be C/m.

Apart from this, I think your approach is correct. Without seeing how you integrated the thing, I can't say more.
 
Okay, yes disregard my question. I was using a computer algebra system (Microsoft Mathematics) to solve the integral. I tried it on my TI-84 Plus and I got .01863 Volts.

(This is what I get for being lazy).
 

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