Solving Gauss's Law Problems: Infinite Sheet & Slab of Charge

[typeinnocent]

Homework Statement

https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2112/summer/homework/Ch-21-Gauss-Law/charged_sheet/sheet.gif
An infinite nonconducting sheet of charge, oriented perpendicular to the x-axis,passes through x = 0. It has area density σ1 = -3 µC/m2. A thick, infinite conducting slab, also oriented perpendicular to the x-axis, occupies the region between x = a and x = b, where a = 2 cm and b = 3 cm. The conducting slab has a net charge per unit area of σ2 = 5 µC/m2.

(a) Calculate the net x-component of the electric field at the following positions: x = -1, 1, 2.5 and 6 cm.
(b) Calculate the surface charge densities on the left-hand (σa) and right-hand (σb) faces of the conducting slab.You may also find it useful to note the relationship between σa and σb.

Homework Equations

Gauss' Law E = σ/2*epsilon

The Attempt at a Solution

PART A
-- For x = -1 cm, I summed the two individual electric fields. For the sheet I said the electric field would be positive since the sheet's charge is negative so the field lines are going in the positive x direction (towards the sheet). For the slab I said the electric field was negative since the slab had positive charge, so the field lines are going away from the slab (i.e. in the negative x direction)
-- For x = 1 cm, same logic. I summed both individual fields again. This time both were negative since field lines for both the sheet and the slab were headed in the negative x direction.
-- For x = 2.5 cm, the electric field is zero since it is within the conductor and conductors have zero electric fields within them.
-- For x = 6 cm, I only used the electric field of the slab, since I thought it would block the field of the sheet. However I said it had a sigma of 2 µC/m2, since 3 µC/m2 had to be on the left side to balance out the -3 µC/m2 of the sheet.

PART B
Since σ = E*epsilon, I multiplied the value of the electric field at 1 cm by epsilon to get the value of σa. I said the charge was positive since it had to counteract the negative area density of the sheet. For σb, I used charge conservation (i.e. 5 - σa).

My question to everything above is: is my logic correct? I really had no idea on how to do the problem so I started taking wild guesses and plugging in numbers. Thankfully all the answers above are right, but come time for a test I want to KNOW how to do the problem, rather than rely on blind luck. Thank in advance for correcting any of my incorrect thinking!

 

[LowlyPion]

-- For x = 6 cm, I only used the electric field of the slab, since I thought it would block the field of the sheet. However I said it had a sigma of 2 µC/m2, since 3 µC/m2 had to be on the left side to balance out the -3 µC/m2 of the sheet.

... My question to everything above is: is my logic correct? I really had no idea on how to do the problem so I started taking wild guesses and plugging in numbers. Thankfully all the answers above are right, but come time for a test I want to KNOW how to do the problem, rather than rely on blind luck. Thank in advance for correcting any of my incorrect thinking!

Conceptually maybe you should probably think more in terms of Gauss Law and "canceling" of charges insofar as you might construct a surface that transects the two planes of charge and are summing what's within. In which case, it's not so much a matter of "blocking" anything so much as merely considering the total charge within your closed surface.

Otherwise, your choice of wording aside, you seem to grasp the material maybe better than you think.