Electrostatics problem inslator and a conductor problem

  • Thread starter koab1mjr
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1. The problem statement, all variables and given/known data
A. Two infinite thin sheets of charge are placed perpendicular to the x axis. The surface charge densities of the two plates are (SC density 1 = 8.5 C/M2) and (SC density 2 = 1.5 C/M2. Placed near plate 2 is a think conducting metal plate with a (SC density of -3.0 C/M) Find the magnitude of the field at the origin

problem has a diagram

d = 60 cm

plate 1 (2d space separation) origin (d space separation) plate of thickness d (d space separation) plate 2

B. what is the surface charge density on the right side of the conducting plate?
2. Relevant equations

The only equation needed is the E field formula for infinte plates which is surface charge density/ 2 * permitivity constant

3. The attempt at a solution
I was applying superposition to this problem realizing that the conductor will be a different formula of surface charge density/ permitivity constant. With this route I got 4.802e11N/C for plate 1 E field, 8.47e10N/C for plate 2's e field and 3.389e11N/C for the conducting plate. Adding them together gave me 2.26e11N/C. The correct answer is 5.65e11N/C. I noticed that if i disregard the plate all together it works out but it might be a coincidence and I do not understand why which is why I am posing the question on the board.

The conducting plate is generating a field and I know the charges will shift to respond to the charged plate to the right of it. But its stronger than that plate and should still be generating a field. Not sure why you would just ignore it any help would be much appreciated.

Since I have been stuck on part a.
no clue on how to break this down but I assume its having the conductor shift charge to keep its center neutral but I am not sure how to go about doing that. I tried calculating the net field experienced by the plate in the center. Not sure if this is correct
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I am assuming my lack of response is for the lack of the equations and stuff

[......] ------2d-------*---d---[.....t........]----d-------[.....]

[tex]\sigma[/tex]1 = 8.5 C/m2
[tex]\sigma[/tex]2 = 1.5 C/m2
[tex]\sigma[/tex]metal plate = -3.0 C/m2
d = 60 cm

Work Formulas

E = [tex]sigma[\tex] / 2 [tex]epsilon[/tex] 0 <-----For Insulating plate
E = [tex]sigma[\tex] / [tex]epsilon[/tex] 0 <----- For Conducting plate

hopefully this is a little better I need to find the field strength at point * and I need to find the [tex]\sigma[/tex]right side of metal plate
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I am sure this problem is within the realm of ability the hw helpers here. Yet no one has answered. If there is an issue with it please let me know what it is and I will try and fix it.

Not sure why no one has answered I have seen post addressed within the hour

what gives!!!

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