How Many Excess Electrons Are on a Charged Rod with Nonuniform Density?

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Homework Help Overview

The discussion revolves around calculating the number of excess electrons on a charged nonconducting rod with both uniform and nonuniform charge density. The rod's dimensions and charge densities are specified, prompting participants to explore the implications of these parameters on the calculations.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculations for both uniform and nonuniform charge densities, examining the formulas used for determining the total charge and the resulting number of electrons. There is a focus on identifying potential errors in the calculations, particularly in part b, where the nonuniform density is applied.

Discussion Status

Some participants have identified discrepancies in the calculations, particularly regarding the order of magnitude and unit conversions. There is ongoing exploration of the assumptions made in the setup and the calculations, with suggestions to verify the units of the constants involved.

Contextual Notes

Participants note that the original poster's calculations for part b may have been affected by unit conversion errors, specifically regarding the conversion from microcoulombs to coulombs. The discussion highlights the importance of careful unit management in physics problems.

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Homework Statement


A charged nonconducting rod, with a length of 1.33 m and a cross-sectional area of 3.40 cm2, lies along the positive side of an x axis with one end at the origin. The volume charge density ρ is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if ρ is (a) uniform, with a value of -4.22 µC/m3, and (b) nonuniform, with a value given by ρ = bx2, where b = -2.66 µC/m5?

Homework Equations


q=ne
p=bx^2

The Attempt at a Solution


So for part a:
n=(p(a)L)/e
n=(-4.22E-6C)(3.4E-4m^2)(1.33m)/(-1.6E-19C)
n=1.193E10 electrons

Part b is where i was incorrect
p=bx^2
p=-2.66x^2
dq=Apdx
=3.4E-4*-2.66x^2dx
so q=9.04E-4* the integral of x^2dx from 0 to 1.33
q=-7.09E-4C
n=q/e
=-7.09E-4C/-1.6E-19C
=4.431E15 electrons<----- this was incorrect and I am unsure of my mistake
 
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Jrlinton said:
Part b is where i was incorrect
p=bx^2
p=-2.66x^2
dq=Apdx
=3.4E-4*-2.66x^2dx
so q=9.04E-4* the integral of x^2dx from 0 to 1.33 ⇐
q=-7.09E-4C
n=q/e
=-7.09E-4C/-1.6E-19C
=4.431E15 electrons<----- this was incorrect and I am unsure of my mistake

Looks like the order of magnitude went astray starting here. 10-4 is too big. Check the units of the constant b.
 
Right, so multiply the final answer by the conversion factor of microcoulombs to coulombs of E-6.
 
That'll work.
 

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