Infinite Sheets of Charge and Conducting Slab

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SUMMARY

This discussion focuses on calculating the potential difference between two points, P and S, in the presence of two infinite sheets of charge and an uncharged conducting slab. The charge densities are σ1 = 0.34 μC/m² for the first sheet and σ2 = -0.14 μC/m² for the second sheet, with the conducting slab positioned between them. The user has successfully calculated the charge densities on either side of the slab as -0.24 μC/m² and 0.1 μC/m², and the potential at point P as 1.492e3 V. The challenge remains in determining the potential at point S, specifically the energy field, which is not constant between points S and P.

PREREQUISITES
  • Understanding of electric fields generated by infinite sheets of charge.
  • Knowledge of potential difference calculations in electrostatics.
  • Familiarity with charge density concepts and their implications on electric fields.
  • Basic principles of conductors in electrostatic equilibrium.
NEXT STEPS
  • Learn how to calculate electric fields due to multiple infinite sheets of charge.
  • Study the concept of electric potential and its relationship with electric fields.
  • Explore the method for determining potential differences in systems with conductors.
  • Investigate the effects of charge distribution on the electric field in the vicinity of conductors.
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Students studying electrostatics, physics educators, and anyone involved in solving problems related to electric fields and potentials in electrostatic systems.

darwin59
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Homework Statement


An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.34 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = -0.14 μC/m2 is located at x = c = 26 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 11 cm and x = 15 cm). What is V(S) - V(P), the potentital difference between point P and point S, located at (x,y) = (20.5 cm, -15 cm)? (point P is located at (5.5 cm, 0 cm))

Homework Equations





The Attempt at a Solution


I've already solved for the left and right side charge density of the conducting slab to be -0.24 μC/m2 and 0.1 μC/m2 respectively. I've also already solved the potential at point P, being 1.492e3 V (or at least I think I have). What I need help with is finding the potential at point S, more specifically the energy field there. The program tells me I'm on the right track but the energy field is not constant from S to P, which I already know, but I can't figure out how to solve for it.
 
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darwin59 said:

Homework Statement


An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge density σ1 = 0.34 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = -0.14 μC/m2 is located at x = c = 26 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 11 cm and x = 15 cm). What is V(S) - V(P), the potential difference between point P and point S, located at (x,y) = (20.5 cm, -15 cm)? (point P is located at (5.5 cm, 0 cm))

Homework Equations



The Attempt at a Solution


I've already solved for the left and right side charge density of the conducting slab to be -0.24 μC/m2 and 0.1 μC/m2 respectively. I've also already solved the potential at point P, being 1.492e3 V (or at least I think I have). What I need help with is finding the potential at point S, more specifically the energy field there. The program tells me I'm on the right track but the energy field is not constant from S to P, which I already know, but I can't figure out how to solve for it.
Hello darwin59. Welcome to PF !

The charge densities on the on either side of the slab have to cancel, i.e., they have to be equal in magnitude & opposite in sign.

The Electric Field between the two sheets of charge and external to the conducting slab is the same as if the slab were not present.
 
Thanks for the hospitality!

I'm not sure how I didn't try that out, or how exactly I got stuck down the path that the conducting slab should have a net charge, but I got it. Thank you very much SammyS!
 
You're welcome.

Most of us try to give "newbies" a warm welcome.
 

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