Recent content by putongren

Using Linear Algebra to solve this problem boggles the mind. Did you use matrices or vectors?

After thinking about the problem along with your helpful input, logically, the problem ends when the trains collide. d/2 = vt -> t = d/(2*v) Since "t" is the total time it takes for the problem to end, multiply “t” with u (bee’s velocity) to find the total distance the bee travels. Therefore...

This is a question from the MIT Open courseware website. (1). d = vt + ut let t = time it takes d = (u + v)t t = d / (u + v) (2). d = vt + ut d - vt = ut. Substitute t with d / (u + v) d - v*(d/(u+v)) = u*(d/(u+v)) d - v*(d/(u+v)) = “distance...

This question is from the MIT Courseware. I’m having difficulty finding the general equation to solve the problem (1). d = vt + ut d = (u + v)t t = d/(u + v) (2). d = vt + ut d - vt = ut sub t with d/(u+v) d - (v*d)/(u+v) = (u*d)/(u+v) I’m done with the...

I guess knowing the definition of an exact differential would help in answering my question: https://en.wikipedia.org/wiki/Exact_differential

I was researching about conservative and non-conservative forces, and there is some information in a website that sates that the work done is independent of path if the infinitesimal work 𝐹⃗ ⋅𝑑𝑟⃗ is an exact differential. It further states that in 2 dimensions the condition for 𝐹⃗ ⋅𝑑𝑟⃗ = Fxdx...

I'm just wondering, the charges are on the surface, and the electric field is inside the sphere. Does the charges qualify as in the same region as the region inside the sphere? The original question states that the charges has to be in the same region as the electric field.

How about a sphere with constant charge density on the surface? The electric field is 0 on the inside of the sphere, but there are charges on the surface.

This is a conceptual question. I think we can conclude that electric charges cannot be present if there is no electric field in that region. Is this an application of Gauss' Law? A net electric flux thru a surface indicates that there is a charge within that region. An electric field must be...

I understood the situation intuitively and conceptually concerning the equilibrium issue, but couldn't wrap my head around how the mathematics dictate whether the equilibrium is stable or not.

Thanks kuruman for the graph. I think the blue curve is the curve that will return the third bead towards the equilibrium. The potential energy in both ends of the curve is positive. This means the third bead will have a tendency to move from positive potential energy to zero, the place of...

I want to use calculus to solve the stability issue. It seems like a more elegant way to do solve the problem. I want to find the first and second derivative of potential as mentioned in post #16. Also, when you mention potential, do you mean kq/r^2? How do we go about finding the first and...

let d = distance between charges x = displacement from equilibrium \begin{equation} \notag a =point of equilibrium = \frac{\sqrt{3}*d}{(1+ \sqrt{3})} \end{equation} then \begin{equation} \notag \frac{k*3q}{(x+a)^2} - \frac{k*q}{(d-(x+a)^2} = Net Field \end{equation} Excuse my poor use of...

I'm not sure if this is right and I'm not sure if I'm understanding everything that has been said by others thus far. I think we should sample what happens when we let a be greater than the equilibrium point and when a be less than the equilibrium point. If there is a positive net force when...

I am somewhat acquainted with calculus, but it has been a while. Let me try to see if the third bead is stable mathematically.Here is a recap of what I did, sort of: E1 = Electric field from left charge at the third bead E2 = Electric field from right charge at the third bead E1 - E2 = Net...