Electric Field Between Two Infinite Sheets of Charge

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SUMMARY

The discussion centers on calculating the potential difference between two points in the presence of two infinite sheets of charge with charge densities σ1 = 0.57 μC/m² and σ2 = -0.39 μC/m², separated by an uncharged infinite conducting slab. The initial calculation yielded a potential difference of 8685.87 V, based on the assumption of a constant electric field. However, feedback highlighted the necessity to consider that the electric field is zero within the conducting slab, which affects the potential difference calculation. The correct approach requires accounting for the regions of varying electric field strength between the sheets and the conducting slab.

PREREQUISITES
  • Understanding of electric fields generated by infinite sheets of charge.
  • Knowledge of electrostatics, particularly the behavior of conductors in electric fields.
  • Familiarity with the concept of electric potential and potential difference.
  • Basic calculus for integrating electric fields over distances.
NEXT STEPS
  • Study the properties of electric fields between multiple charged plates.
  • Learn about the behavior of electric fields within conductors and the implications for potential difference.
  • Explore the mathematical derivation of electric potential from electric fields.
  • Investigate the effects of charge density variations on electric field strength.
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BRuss9807
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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 = 0.57 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = -0.39 μC/m2 is located at x = c = 28.0 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 12.0 cm and x = 16.0 cm). What is V(S) - V(P), the potentital difference between point P, located at (x,y) = (6.0 cm, 0.0 cm) and S, located at (x,y) = (22.0 cm, -16.0 cm)?

https://www.smartphysics.com/Content/Media/Images/EM/06/h6_planeA.png

Homework Equations





The Attempt at a Solution


I have attempted at an answer and obtained the result of 8685.87 V.

This answer came by way of:

Electric field at point P:
E = (σ1/2ε)-(σ2/2ε) = 54286.72 (N/C) or (V/m)
Set this point to be the zero potential energy position, and since the electrical field is constant, or so I thought, in the x direction, with no component in the y direction, I simply took the difference in x positions of the two points. 22cm-6cm= 16cm, therefore the ΔV between these two points is (54286.72V/m)*(.016m)= 8685.87 V

The feedback to this answer says that I need to account for the Electric field not being constant between these two points. What I am confused about is that I thought the electric field between these two points would be constant. Since a single infinite plate will have a constant electric field that is not dependent on distance from the plate I assumed that the electric field between two of these plates would also be constant. Thank you to anyone that can explain this to me, I can't seem to figure it out and its driving me crazy.
 
Last edited by a moderator:
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BRuss9807 said:

Homework Statement


An infinite sheet of charge is located in the y-z plane at x = 0 and has uniform charge denisity σ1 = 0.57 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = -0.39 μC/m2 is located at x = c = 28.0 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 12.0 cm and x = 16.0 cm). What is V(S) - V(P), the potentital difference between point P, located at (x,y) = (6.0 cm, 0.0 cm) and S, located at (x,y) = (22.0 cm, -16.0 cm)?

https://www.smartphysics.com/Content/Media/Images/EM/06/h6_planeA.png

Homework Equations



The Attempt at a Solution


I have attempted at an answer and obtained the result of 8685.87 V.

This answer came by way of:

Electric field at point P:
E = (σ1/2ε)-(σ2/2ε) = 54286.72 (N/C) or (V/m)
Set this point to be the zero potential energy position, and since the electrical field is constant, or so I thought, in the x direction, with no component in the y direction, I simply took the difference in x positions of the two points. 22cm-6cm= 16cm, therefore the ΔV between these two points is (54286.72V/m)*(.016m)= 8685.87 V

The feedback to this answer says that I need to account for the Electric field not being constant between these two points. What I am confused about is that I thought the electric field between these two points would be constant. Since a single infinite plate will have a constant electric field that is not dependent on distance from the plate I assumed that the electric field between two of these plates would also be constant. Thank you to anyone that can explain this to me, I can't seem to figure it out and its driving me crazy.
What is the electric field between x = 12.0 cm and x = 16.0 cm , i.e., inside the conducting slab ?
 
Last edited by a moderator:
Wow thank you so much SammyS. I cannot believe I forgot that the area inside the conductor would not contain an electric field, therefore no contribute to the potential change. True testament to the need to simply think the problem through completely. Thanks again.
 

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