Recent content by BuggyWungos

:bow: Thank you that's such a silly oversight, I even wrote it as r hat in latex and missed it lol

I'm surprised you were able to understand all that before I fixed it. I understand the second part of your comment, but where is ##\sqrt{x^2+y^2}## in the second step of my prof's solution? Edit: I would have expected it to end up like ##d\vec{B}(r) = \dfrac{\mu_0}{4\pi}...

My attempt: $$d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}r\sin{\theta}}{r^2}$$ $$d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}\sin{\theta}}{r}$$ $$ \sin{\theta} = \dfrac{y}{(x^2+y^2)^{1/2}}$$ $$ d\vec{B}(r) = \dfrac{\mu_0}{4\pi}\dfrac{Id\vec{l}}{r}\dfrac{y}{(x^2+y^2)^{1/2}}$$...

Right, cause ##R=0## and ##\dfrac{\delta V}{R}## would just be infinity :D To finish it off, in my original example, ##\dfrac{\delta V}{R_A} = I_A## and ##I_A## should be equal to the current running through the wires above and at n-m? If that's true, could I also say that the circuits current...

I'm going to simplify my model a little bit as I feel like I'm going in circles trying to understand what exactly is going on. 1) For an example like the one below, if I had thrown on a resistor with resistence any real number, the current through the resistor should be the same current...

Well, if that's true, why doesn't current get smaller as it goes through Resistor A compared to the wire above? If you applied Ohm's law to the wire above, you'd get a much larger current than applying ohms law on the resistor, in fact it would be infinite current... I know something isn't quite...

Say we doubled the resistence in the resistor A, wouldn't the formula spit out current to be half that of what it originallly was due to ##\delta V## remaining constant? $$R_2 = 2R_1$$ $$\dfrac{\delta V}{2R_1} = \dfrac{1}{2}I$$

I understand that the current in the top wire above A will be the same as the wire just below A, as it will be for the wire region marked by n to m, due to the fact that all current in equals all current out. My confusion lies in the first equation I supplied, the change in voltage through the...

In that case, would it be better to imagine a point charge in an electric field to be inbetween 2 oppositely charged plates? I'm still struggling to wrap my head around electric fields a little bit. A lot of diagrams show electric field lines in negative charges to be radially inwards...

I think I understand, it's got to do with how the electric force relates to the field. If the electric force points opposite of the electric field on a point charge, and the point charge is moved alongside the electric force, we wouldn't really need to do any work (negative work) as the electric...

My understanding of this question is that, if you have a proton standing against a positive electric field, and move it in the opposite direction of the field, you're putting in work and therefore should have greater electric potential energy. But that idea breaks down when you consider a...

What is the correct electric field strength formula using rho? I understand that E = rho/(epsilon nought) would give the correct answer, but the formula I was given in my textbook was E = 2(pi)k(rho), which would simplify to E = rho/2(epsilon nought). Is the above formula used for another situation?

surfafce area = 0.502 E = -q/A2(en) = 3800 -q = 3800*(A2(en)) -q = 1.68*10^(-8) -q = 3.37*10^(-8) V = kq/r V = (9.0*10^9)(-3.37*10^(-8))/0.2 V = -1519 V