
#1
Feb413, 12:23 PM

P: 135

1. The problem statement, all variables and given/known data
An infinite sheet of charge is located in the yz plane at x = 0 and has uniform charge denisity σ1 = 0.3 μC/m2. Another infinite sheet of charge with uniform charge density σ2 = 0.33 μC/m2 is located at x = c = 21.0 cm.. An uncharged infinite conducting slab is placed halfway in between these sheets ( i.e., between x = 8.5 cm and x = 12.5 cm). What is V(S)  V(P), the potential difference between point P and point S, located at (x,y) = (16.75 cm, 12.5 cm)? (point P is at 4.25 cm, 0 cm) 2. Relevant equations no quite sure 3. The attempt at a solution So, i thought that the electric field would be constant between the two plates. But, apparently it's not (which i still don't fully understand). Also, i realize that i can basically ignore the thick slab since it has a net of 0 charge. I got v(p) to be 1512.71, but i do not know how to go about getting v(s). Or rather the e field at s would be a good start. I know the e field at p is 35593.2. 



#2
Feb413, 01:06 PM

HW Helper
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PF Gold
P: 4,481

Hello, Gee Wiz. Welcome to PF!
Have you learned about electric fields inside of conducting materials for electrostatic conditions? 



#3
Feb413, 01:07 PM

P: 135

um...you mean like induced charge? (i thought that was for spheres)..or did i miss your question?




#4
Feb413, 01:13 PM

HW Helper
Thanks ∞
PF Gold
P: 4,481

Potential of Infinite Sheets of Charge and Conducting Slab
Have you learned about the magnitude of E inside a conductor for electrostatics? There's a general fact about this that applies to all conductors.




#5
Feb413, 01:15 PM

P: 135

Oh, i think i know what you are referring to. That the e field within a conductor's metal is 0. ya?




#6
Feb413, 01:18 PM

HW Helper
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PF Gold
P: 4,481

ya ya




#7
Feb413, 01:20 PM

P: 135

alright, i'm not quite sure how that helps here. I thought i could basically just ignore that slab since it has a net of 0




#8
Feb413, 01:22 PM

HW Helper
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PF Gold
P: 4,481

If you didn't have the slab, how would you get the potential difference?




#9
Feb413, 01:23 PM

P: 135

take the efield that i found at p times the difference between the distance of the two points.




#10
Feb413, 01:29 PM

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PF Gold
P: 4,481

When you take the distance between the two points, be careful. Do you take the actual displacement between the points, or only the component of the displacement parallel to the field?
With the slab in place, there is no electric field in the region of the slab. Should the potential change when going through the slab? 



#11
Feb413, 01:32 PM

P: 135

Sorry, when i said distance between the two points i meant horizontal distance.
I think i am struggling with the slab notion. I'm pretty sure that the efield and therefore the potential charge in the slab would be zero. I found in an earlier part of this problem the charge density on either side of the slab (.315uc/m^2 and .315uc/m^2). Does that play a role here? 



#12
Feb413, 01:43 PM

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PF Gold
P: 4,481

Suppose that you had two points of space where there is no electric field in the region between the points. How would the potential at the two points compare? For example if V = 100 V at one point, what would it be at the other point? Can you apply this to two points of the slab? Say, point A on the left surface of the slab at x = 8.5 cm and point B on the right surface at x = 12.5. What is V(A)  V(B)? 



#13
Feb413, 01:47 PM

P: 135

i think i meant electric potential (since e field is zero).
If there is not electric field in the region then i would assume that the other point would also be 100V. I think? V(A)V(B) would be 0 since there is no e field in the slab..? 



#14
Feb413, 01:52 PM

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PF Gold
P: 4,481

That's right. The potential has the same value for all points of the slab.




#15
Feb413, 01:57 PM

P: 135

Alright so, would i just shrink the distance (horizontal distance) by 4cm. since that is the length of the slab. so instead of being 16.75cm away from the furthest charge it is only 12.75cm away?




#16
Feb413, 02:00 PM

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PF Gold
P: 4,481

Yes. Or, equivalently, shrink the horizontal distance between P and S by 4 cm.




#17
Feb413, 02:03 PM

P: 135

Awesome, two things: i got the right answer from that and your explanation was really helpful for other things in my course. I really appreciate it. Have a super Monday!




#18
Feb413, 02:11 PM

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PF Gold
P: 4,481

Good work!



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