Calculating Work Done by Electric Field on a Point Charge

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Homework Help Overview

The problem involves calculating the work done by an electric field on a point charge moving in the vicinity of a long charged line with a specified negative linear charge density. The context is rooted in electrostatics, specifically dealing with electric fields and forces acting on charges.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the provided electric field equation and Gauss' Law, with one attempting to derive the work done by integrating the force over the distance. There is also confusion regarding the integration process and the meaning of linear charge density.

Discussion Status

The discussion is active, with participants exploring different interpretations of the equations involved and questioning the integration steps. Some guidance has been provided regarding the setup of the integral for work, but there is no consensus on the integration method or the meaning of specific terms.

Contextual Notes

There is a lack of clarity regarding the term "wavelength," which appears to be a misunderstanding, as it relates to linear charge density in this context. Participants are also navigating the implications of the charge density and its role in the calculations.

05holtel
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Homework Statement




Consider the electric field created by a very long charged line of negative linear charge density -2.50 nC/m.

A small positive point charge of 8 mC moves from a distance of 9 cm to a distance of 17 cm.

How much work is done by the electric field?

Hint: The electric field for a long charged line is:

The Equation is E line = 1/(4piE0)(2lamda/r)


Express the result in the unit mJ and to three significant figures.

Homework Equations




The Equation is E line = 1/(4piE0)(2lamda/r)

The Attempt at a Solution



Is this hint equation provided equivalent to Gauss' Law: (2*pi*R*L*E = L*q/eo)

where eo is the coulomb's law constant "epsilon-zero" = 8.89^10^-12 N*m^2/C^2
and q is the charge per unit length

Therefore,

E(R) = q/(2 pi eo R)
Multiply that by Q = 8 mC and integrate with dR from R = 0.09 to 0.17 m

Also I am confused how to integrate this
 
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E = \frac{\lambda}{2 \pi \epsilon_o r}
W = \int\limits_{R_1}^{R_2} F_e\, dr
W = \int\limits_{R_1}^{R_2} qE\, dr
W = \int\limits_{R_1}^{R_2}\frac{q\lambda}{2 \pi \epsilon_o r}\, dr
Everything is constant except for the r.
W = \frac{q\lambda}{2 \pi \epsilon_o }\int\limits_{R_1}^{R_2}\frac{1}{r}\, dr
W = \frac{q\lambda}{2 \pi \epsilon_o} ln(r)|\limits_{R_1}^{R_2}
 
Thanks so much,
Do you know what the wavelength is
 
05holtel said:
Thanks so much,
Do you know what the wavelength is

\lambda is linear charge density
 

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