Recent content by ab200

I see that in 1), the two charges are symmetric about the y axis. However, in case 2) I’m a little bit confused why they cancel each other out. Yes, they are equal distances from the origin, but does that necessarily mean they cancel out? If the bottom middle and top right charges were switched...

Yes, I figured this out. I was missing that the first part of the question actually asks for electric potential V rather than potential energy U. The equation for the electric potential of the system (multiple) of point charges is V = V1 + V2 + V3. You’re right about the second part as well. I...

Yes, in this case work is equal to the change in potential energy. The electron charge for part 2 of the question would be q4, with the relevant U values being U14, U24, and U34.

Looking at the image, I see that due to symmetry, the bottom-left negative charge and the bottom-right positive charge cancel out, leaving me with a triangle around the center. I'm not entirely sure how to solve for potential at the origin specifically, but I believe that the potential energy of...

I see what you mean and why that is the goal, but I‘m still short-circuiting on what to do in order to get there. For instance, I suppose I could say θm = sin-1[(mλ)/d] = tan-1(ym/L), but is that really a sufficient symbolic expression? And assuming that it is, how does that help me find θo...

This drawing is from my notes, but this is also what I was thinking for this problem. (Ignore the s1, that’s just a reference point.) I‘m still a bit confused by what you mean with “the condition for each max assuming β to be held fixed.” In this case, β is explicitly held fixed at 40 degrees...

Alas, it is not. I figure it out though! Thank you for your help.

I got it! Thanks for the tip.

Given that [y][/m] is equal to the position of the interference maxima and is the variable I’m solving for. I got as far as setting up two expressions, one for sinθ and the other for tanθ. sinθ = (mλ)/d tanθ = ym/L My first instinct was to set θ = 40 degrees. By relating sinθ and tanθ to each...

Putting the questionable nature of my professor's problem-writing aside, I found this question pretty confusing. I assumed that 60 m was the focal length, and that R1 was infinity since it is a flat surface. This gives me: 1/60 = (1.5 - 1)[(1/∞) - (1/R2)] This gave me R2 = 30 m, but that isn't...

Since the lens is convex, I figured that the points where the red and blue light focus on the optical axis would be equal to their respective focal lengths (f), given that the incoming rays are parallel to each other and perpendicular to the lens. Solving this got me to 1/fred = (1.52 -...

I apologize — after working on it again I arrived at the correct answer. I’m not entirely sure what I did the first time, but I suspect I didn’t convert degrees into radians when calculating difference in initial phase.

The first part of this question asks for the maximum amplitude produced, which I found to be 12.86 Pa. I was able to set up the expression for the combined wave equations. However, I am struggling to understand how to set up and solve an expression to find the distance I could move one of the...