Regarding Question 1:

For part A.), has Gauss's law of electric flux been discussed at all? Particularly, the method of closed Gaussian surfaces for calculating electric fields? Here are a few questions that may help:

(1) For a negative charge distribution as given, what direction does the electric field point to on either side of the sheet? Does it point toward or away from the charge?

(2) Does the electric field make an angle to the sheet? Why or why not?

Consider the symmetries. We can rotate the sheet in its own plane (around the x axis) by any angle without fundamentally changing the problem: it would still be an infinite sheet of charge. Similarly, we can flip it by 180°, mirror it side-to-side or slide it around in its own plane (y-z). Neither of these operations change the problem, so they also can not change the solution.

Now picture the electric field moving and rotating with the sheet as we perform these operations. Wherever the electric field vectors end up at, they better be indistinguishable from the field before the operation (or it won't be the same solution)!

What kind of fields directions are allowed and what kind are forbidden because of the symmetry?

(3) If you can answer the questions above, you should have a basic idea of what the electric field looks like. Now you can use a Gaussian surface to find the electric field strength. Gauss's law says that:

The total electric flux (flux = the electric field perpendicular to the surface × the area) through any closed surface is equal to the total charge divided by ε0 inside that surface:

∫ E ∙ n dA = Q/ε0

So if you can find a shape where the strength electric field and total flux are related in some easy way (For example, because the field is parallel to some sides and contributes no flux there and/or perfectly perpendicular to other sides where you only have to divide by the area) you can calculate the field strength just from the total charge.

Can you come up with such a shape?

(4) Now, how much charge is inside that shape? And therefore, how much should the total flux be (Gauss's law)?

(5) How is that flux divided over all the sides? How large is the perpendicular electric field on those sides (use definition of electric flux)?

(6) Now, what is Ex on either side of the sheet (note direction: + for field pointing right, − for left)? Does it increase, decrease or stay the same if you go further out from the sheet?

If you can answer all that, you should be able to master Gaussian surfaces to calculate the electric field around any distribution of charge.

Now for part B.) the electric potential increases if you move against/into the direction of the electric field, and decreases if you move with it. The amount of change is directly proportional to the strength of the field.

(1) So as you move along x, does the potential increase (moving against the field) or decrease (moving with the field)?

(2) Does the proportion by which it is increasing/decreasing change (because the electric field strength changes)? If the proportion is constant, what kind of graph does that make?

(3) Electric potential is continuous, so make sure that both sides of the sheet connect at the same potential.

Regarding Question 2:

A.) Yes, the potential in point A is around 5V (it is on the 5V line). The potential in point B is halfway between 7V and 8V. So point B has the higher potential with 7.5V.

B.) Correct. As with 1B.) the electric field strength is proportional to the change of potential. A rapid change of potential (lines closer together) means a stronger electric field.

C.) Correctly drawn. The electric field points in the direction of decreasing potential. And from B.) you already concluded that the field in point A was stronger.

D.) First: what is the change in potential (in Volts) from point C to point A?

Now multiply that by the charge on the proton (which is the same as the charge on the electron, but +e rather than −e) to get the change in potential energy. Hint: 1e × 1V = 1eV (which is the electronvolt, a small unit of energy). If you need a numerical value as answer, 1eV = 1.6×10⁻¹⁹ J.

Whew.. this became a bit longer than I thought, sorry about that. Still, I think if you work trough the questions I posed (you don't need to answer them all, just for your guidance), you should be able to master this part of the exam.