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Infinite Sheets of Charge and Conducting Slab

  • Thread starter darwin59
  • Start date
12
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1. 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))
2. Homework Equations



3. 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.
 

SammyS

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1. 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))
2. Homework Equations

3. 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.
 
12
0
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!
 

SammyS

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Homework Helper
Gold Member
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924
You're welcome.

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

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