Infinite Sheets of Charge and Conducting Slab

[darwin59]

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.

 

[SammyS]

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.

 

[darwin59]

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]

You're welcome.

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

 

[asteriatic]

I would first like to commend you on your progress in solving this problem. It seems like you have a good understanding of the concepts involved and have made significant progress in finding the potential at point P and the charge density of the conducting slab.

To solve for the potential at point S, we can use the formula for the potential due to an infinite sheet of charge, which is V = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space. Since point S is located at (x,y) = (20.5 cm, -15 cm), it is closer to the negative sheet of charge (located at x = 26 cm) than the positive sheet of charge (located at x = 0). Therefore, we can use the surface charge density of the negative sheet (σ2 = -0.14 μC/m2) to find the potential at point S.

V(S) = σ2/2ε0 = (-0.14 μC/m2)/2ε0 = -7.0e3 V/m

Now, to find the potential difference between points S and P, we can use the formula V(S) - V(P). Plugging in the values, we get:

V(S) - V(P) = (-7.0e3 V/m) - (1.492e3 V) = -8.492e3 V

Therefore, the potential difference between points S and P is -8.492e3 V, meaning that point S is at a lower potential than point P. This makes sense, since point S is closer to the negative sheet of charge, which has a lower potential than the positive sheet of charge.

I hope this helps in solving the problem. Keep up the good work!