# Recent content by Harrisonized

1. ### Magnetic Monopole (Magnetic Charge)

I saw that before I made this thread. That's not quite as much information as I'd hoped. I was hoping for sort of like an introduction on the subject. For example, the first textbook I ever learned E/M out of is by Arthur Kip. Each chapter is organized in two parts. First, they present a law...
2. ### Batteries connected in parallel

Consider the following circuit: http://www.ibiblio.org/kuphaldt/electricCircuits/DC/00207.png The voltage at the node between the resistors is: V = (V1/R1+V2/R3)/(1/R1+1/R3+1/R2) Suppose R1 = R3 = R. Then: V = (1/R)(V1+V2)/(2/R+1/R2) V = (V1+V2)/(2+R/R2) Now let R go to zero...
3. ### Magnetic Monopole (Magnetic Charge)

An electric charge produces a Coulomb electric field: E = dqe r/r3 A current element produces a Biot-Savart magnetic field B = i dl×r /r3 From what I understand, magnetic charges are inserted for the sake of making Maxwell's equations symmetric. A magnetic charge is meant to produce a...
4. ### Integral of Electric Field is Zero

Hi everyone. This isn't a homework problem. Rather, I'm trying to understand how the δ term arises from the field of a dipole. Homework Statement Greiner supplies the following one-line derivation, which is easy to follow I guess, but doesn't make logical sense to me. Specifically, I don't...
5. ### Fractional Calculus and Residues

According to Fractional Calculus, the power rule can be written as (dm/dzm) zn = n!/(n-m)! zn-m For example, (d1/2/dz1/2) z1/2 = (1/2)!/(1/2-1/2)! z0 = (1/2)√π To find the residue of f(z) = f(z)/(z-z0)m at z→z0, the formula is Res(z→z0) f(z) = 1/(m-1)! dm-1/dzm-1 f(z). For...
6. ### Passive Circuit Elements

What do you mean am I sure? Does that formula look incorrect to you or something?
7. ### Passive Circuit Elements

The three circuit elements are capacitors, resistors, and inductors, which act in the following manner: Capacitor: V = (1/C) q Resistor: V = R dq/dt Inductor: V = L d2q/dt2 Is it possible to build a passive device that acts like: V = (const.) d3q/dt3 Google search has come up with...
8. ### Curl of A

Simple question. It came out of lecture, so it's not homework or anything. My professor said that the curl of a vector field is always perpendicular to itself. The example he gave is that the magnetic vector potential A is always perpendicular to the direction of the magnetic field B. (I haven't...
9. ### I don't understand why it's chiral. It's not like any of the phenyl

I don't understand why it's chiral. It's not like any of the phenyl rings are stuck in a single resonance structure.
10. ### Best Organic Chemistry Book?

Not that I'm trying to necropost, but I finally obtained a copy of this book. It is amazing, and suits all of what I asked for.
11. ### How to get 100% in Physics course?

Head down to your university library, and find the section where they store all the old editions of physics textbooks. Sit down, and read through the section you're learning, and pick out two that you think explains that section best. This is important. First, this teaches you how to judge the...
12. ### Calculators Vector capable calculator recommendations?

Just use your TI-83. Dot products and cross products can both be expressed by matrices, which is included in the TI-83. Alternatively, you can write a program that just spits out your desired numbers.
13. ### Charges are connected and then kicked by chuck norris

Well, here's how I would have done it. Fi = kq1q2r-2 Ff = kqf2r-2 ↓ qf = 0.5(q1+q2) = 0.25k(q1+q2)2r-2 q2 = (4Ff k-1r2)1/2 -q1 Ff = 0.25k(q12+2q1q2+q22)r-2 = 0.25k(q12+q22)r-2+0.5Fi 4(Ff -0.5Fi)k-1r2 = q12+q22 4(Ff -0.5Fi)k-1r2 = q12+((4Ff k-1r2)1/2 -q1)2 4(Ff -0.5Fi)k-1r2 = q12+(q12-2(4Ff...
14. ### Charges are connected and then kicked by chuck norris

Because of this?
15. ### A balloon lifting a heavy stone

Your method for part 3 is correct. However, your method for part 1 is incorrect, which led to an incorrect answer. Next time, try to do the problem with variables as opposed to numbers. It looks cleaner, and it's easier to identify where you went wrong. In the setup of the problem, the balloon...