Calculating Work Done by Electric Field on a Point Charge

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SUMMARY

The discussion focuses on calculating the work done by an electric field on a point charge moving in the vicinity of a long charged line with a negative linear charge density of -2.50 nC/m. The relevant equation for the electric field is E line = 1/(4πE0)(2λ/r), where E0 is the permittivity of free space (8.89 x 10^-12 N*m²/C²) and λ is the linear charge density. The work done, W, is calculated using the integral W = ∫(qE) dr, which simplifies to W = (qλ/(2πE0)) ln(r)| from R1 to R2. The final result is expressed in millijoules (mJ) to three significant figures.

PREREQUISITES
  • Understanding of electric fields and point charges
  • Familiarity with Gauss' Law and its applications
  • Knowledge of calculus, specifically integration techniques
  • Basic concepts of linear charge density
NEXT STEPS
  • Study the application of Gauss' Law in electrostatics
  • Learn about the integration of functions in physics contexts
  • Explore the concept of electric potential energy in electric fields
  • Review the relationship between electric field strength and work done
USEFUL FOR

This discussion is beneficial for physics students, electrical engineers, and anyone studying electrostatics or working with electric fields and point charges.

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
 
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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