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.

 

[nlightner]

Your approach to solving this problem is mostly correct. However, there are a few things that could be improved upon.

First, when calculating the electric field due to the sheet of charge, you should use the formula E = σ/2ε instead of E = σε. The latter is missing a factor of 2 in the denominator, which will result in incorrect values for the electric field.

Second, when calculating the electric field at x = -1 cm and x = 1 cm, you should take into account the fact that the electric field due to the sheet of charge will be constant at all points along the x-axis. This means that the electric field due to the sheet will be the same at both x = -1 cm and x = 1 cm, and you can simply multiply this value by 2 to get the total electric field at these points.

Third, for x = 6 cm, you are correct in saying that the electric field due to the sheet will be blocked by the conducting slab. However, the value of the electric field due to the slab will not be 2 µC/m2, as you have calculated. Instead, you should use the formula E = σ/ε to calculate the electric field due to the slab, since it is a conductor. This will give you a value of 5 µC/m2, which you can use to calculate the total electric field at x = 6 cm.

Finally, for part B, you are correct in using the formula σ = Eε to calculate the surface charge density on the left-hand and right-hand faces of the conducting slab. However, you should use the total electric field at x = 1 cm (which you calculated correctly in part A) to calculate the surface charge density on the left-hand face, and the electric field due to the slab (which you calculated correctly in part A) to calculate the surface charge density on the right-hand face. This will give you values of σa = -2.5 µC/m2 and σb = 7.5 µC/m2.

In conclusion, your approach to solving this problem is mostly correct, but there are a few minor errors that should be corrected. Overall, your understanding of Gauss's Law and electric fields is good. Keep practicing and you will become more confident in solving these types of problems.