# Tag Info

24

Actually, we do! It's just that it's not the same "kind" of potential - and the reason for this is that magnetic forces work differently than electric forces. Magnetic fields, if you know, do not directly exert a force on charged particles, simply for being charged. (They would exert such on hypothetical "magnetically charged particles", but we've never ...

12

The potential is a kind of primitive function of a vector field, primitive in the sense of being the reverse of a differentiation, ie., an integral with a variable upper limit. The derivatives of the potential in all directions represent the vector field; not all vector fields can be represented that way but some do, for example the electrostatic field. Some ...

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There is a "magnetic potential" that appears in more advanced books and is defined as ${\bf B}= \nabla \phi$, just as ${\bf E}= -\nabla V$. (please excuse my equations if you are not familar with the $\nabla$ symbol) It's useful in regions where there is no current or no time-changing electric field, but it is "multivalued." Because of Ampere's law, if you ...

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An eV Electron volt is equal to a volt This cannot be, as the election volt is a unit of energy, whereas a volt is a unit of potential (energy per unit charge). However, they are closely related.... So that volt it is equal to, is it based on 1 volt rms or one volt peak to peak or one volt average, or some other calculation of a volt? The electron volt ...

4

$V$ in ohms law is the difference between potentials, by definition. Now, why would the law require subtraction instead of addition? Imagine than instead of voltage your law is proportional to the length of some object. If in you system of coordinates, one end of the object has a coordinate $x_1$ and the other end a coordinate $x_2$, the length would be the ...

3

(a) There's no ambiguity in this form of the equation $$V_a -V_b= -\int_b ^a \mathbf E\cdot \text d \mathbf l$$ Notes (i) We could swap a and b (on both sides of the equation). In fact I'd prefer them swapped, but that's only a matter of taste. We'll leave them as they are in what follows. (ii) I think it's usually understood in this context that $V_{ab}... 2 FWIW, $$\lim_{\varepsilon\to0^+}\frac{3}{4\pi}\frac{\varepsilon}{(r^2+\varepsilon)^{5/2}}~=~\delta^3(\vec{r})$$ is a well-known representation of the 3D Dirac delta distribution. It is straightforward to derive using test functions, cf. e.g. my related Phys.SE answer here. 2 ...if acceleration is$0$it means that the external force is$0$... No. If acceleration is$0$then net force is$0$. ...and hence net work done should always be zero. Yes, in this scenario the net work is in fact$0$, since the net force is$0$. However, this means that there are (at least) two forces acting on the object in question: gravity$F_g$... 2 It's completely arbitrary. Just choose a direction you want. After using Kirchhoff's voltage law and Kirchoff's current law, if current becomes negative, that'd mean direction of current is opposite, else your primary choice is correct. However, if you don't know Kirchhoff laws, you can assume that the direction of current will be determined by the most ... 2 There is the vector potential$\bf A$for which${\bf \nabla} \times {\bf A} = {\bf B}$. So$B_z = dA_x/dy - dA_y/dx$and similar for other components. The classical field theory of electromagnetism is based on the four potential$A^\mu = (\phi,{\bf A})$.$\phi$is the Coulomb potential . 2 A charged particle in a magnetic field maintains its kinetic energy. You can see this from the fact that$\vec{F}_B \propto \vec{v} \times \vec{B}$which means that$\vec{v}\cdot\vec{F}_B=0$and therefore the force by the magnetic field cannot preform any work on the particle $$W_B = \int \vec{F}_B \cdot \vec{dx} = \int dt \vec{F}_B\cdot\vec{v} = 0$$ This ... 2 I guess we can try using semi-classical approximation. Notice that in one 'period' the particle moves from$x=0$to$x=a$and back. This gives a phase shift$\gamma = -\pi$since there are two smooth turning points in such an orbit. Use the Bohr-Sommerfeld equation ($n \in \mathbb{N} = \{0,1,2,...\}$): $$\frac{1}{\hbar}\oint p_x dx = 2\pi n - \gamma = 2\pi(... 2 The integral$$\int_\gamma \mathbf{E} \cdot d\mathbf{r}$$represents the work done in moving a unit of charge through the electric field along the path \gamma. Note this has just the same form as the work integral$$W_\gamma = \int_\gamma \mathbf{F} \cdot d\mathbf{r}$$only we're using electric field (in effect, force per unit of electric charge ... 1 Thanks to the comments below; my doubt was due to a misunderstanding of notation that Griffiths uses in the book. The right way to look at the first formula is$$ V(\vec{r}) = \frac{1}{4\pi\epsilon_0}\int\frac{\rho(\vec{r}')}{|r-r'|}d\tau' $$From here it is possible to derive the multipole expansion. 1 Usually we indicate the potential difference with an arrow pointing against the direction of the electric field, i.e. where the field decreases, there the potential increases. Potential difference (voltage) is not usually shown as an arrow. Usually you label one terminal positive and one terminal minus. The assignment plus to minus is by convention in ... 1 I understand your struggle with the notation. Unfortunately many texts and teachers us the letter V for both voltage and electrical potential in the same context:$$V_{ab}=V_a-V_b.$$If we ask someone to put this equation in words one might say "the voltage across the space from location$b$to location$a$is the difference in the electrical potentials ... 1 First of all, I assume that in the second term in the "right" answer 2 should be in denominator; if this is the case, it actually satisfies the equation necessary (as$\frac{\partial^2}{\partial z^2} |z| = 2 \delta(z)$in terms of distributions) Both your answer and the other one are actually OK; different gauge transformations can lead to Couloumb gauge. ... 1 An electron volt is not equal to a volt: an electron volt is the change in energy you get by moving a charge equal to that on an electron through a potential difference of one volt. An electron volt is a unit of energy. You can then work backwards through the definition of a volt, which is the potential difference across a conductor such that a current of ... 1 Note that the notion of gauge theories is much more general than, say, Yang-Mills theory. There is of course a long list of gauge theories that have gauge potentials -- e.g. in the case of Yang-Mills theory, the$A^a_{\mu}(x)\$ field is the gauge potential -- but it is not a general requirement for a gauge theory.

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If there is no resistance charge will accelerate in an electric field. In an ohmic resistor current saturates and is constant as you say correctly. The excess kinetic energy is converted into heat by friction. The current distributes in such a way that the potential is approximately constant over a cross section of the resistor. Because of current ...

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