Hi,
Adiabatic, iosbaric and isochoric processes are good approximations for a lot of thermodynamic phenomena in every day life.
But the conditions for a process to be isothermal are so artificial that i have grave difficulties to fudge a story.
Are there any examples of thermal...
Thank you for your fast and detailed answer! That supports my assumption that several authors leave the reader deliberatly (or not) behind some serious issues :uhh:
Now my thoughts about this are about to come to maturity...
Any other suggestions are appreciated as well =)
Hi,
isn't it a bit dangerous to claim that
\left[ x \cdot \left( \psi(x,t) \, \frac{\partial \psi^\ast (x,t)}{\partial x} + \psi^\ast(x,t) \, \frac{\partial \psi(x,t)}{\partial x} \right) \right]_{x=-\infty}^{x=\infty} = 0
for example?
Expressions like this one are often found in...
Biot-Savart's law can handle open ends, but there exists no vector potential for non closed currents (since it's not physical). For non closed currents the vector potential diverges.
For such an unsymmetric problem, where no closed current flow exists it would be the best to solve Biot-Savart's equation for discrete points (if possible) or numerically. I don't think, that a analytical solution exists!?
In addition i don't want to waste hours/days to solve an integral...
Hi,
i like the problems/exercises in the book
Physics for Scientists and Engineers, Lawrence S. Lerner, Jones & Bartlett Publishers
i think they teach good the physics behind the exercises, though only the solutions of the odd numbered problems are given. In addition they only take calculus...
Another possibility to convince you: how much is the acceleration of an object if the energy flow (Power) on it is constant? (as one dimensional treatment)
P = v(t) F(t) = m v(t) a(t) = m v(t) \dot v(t) = \frac{m}{2} \frac{\mathrm d}{\mathrm dt} \Bigl( v^2(t) \Bigr)
integration of both sides...
The connection between the force exerted on an object and the power which is dissipated is
P = \frac{\mathrm dW}{\mathrm dt} = \frac{\mathrm d}{\mathrm dt} \Bigl( \int \mathrm dt ~ \frac{\mathrm d\vec r(t)}{\mathrm dt} ~ F(\vec r,t) \Bigr) = \vec v \cdot \vec F
Maybe it is helpful for you to...
i guess you have the average power in mind!? The power the motor must provide at any time is
P = \frac{\mathrm dW}{\mathrm dt} \qquad \mbox{with} \qquad W = \int \mathrm dr ~ F = \int \mathrm dt ~ v(t) \, F
(i treat the problem as one dimensional) so
P = \frac{\mathrm dW}{\mathrm dt} = \int...
Maybe you know that the electric field is crucial for your consideration!?
I know that's slightly abstract, but the energy is not stored in the capacitor itself. It is stored in the electric field (only the sources and sinks of the electric field are located on the capacitor's plates).
In...
Are you used to the formula of the energy stored in an electirc field
W = \frac{\varepsilon_0}{2} \int \limits_\mathcal{V} \mathrm dr^3 ~ \vec E^{\, 2}(\vec r)
?
you're right
\hat r = \frac{\vec r}{|\vec r^{\,}|} \qquad \Leftrightarrow \qquad \hat r \cdot |\vec r^{\,}| = \vec r
this applied to the primary equation yields
\phi (\vec r) = \frac{1}{4\pi\varepsilon_o} \frac{\vec p \cdot |\vec r^{\,}|\hat{r} }{r^3} =...
You might expand the fraction with |\vec r|!
This yields
\phi (\vec r) = \frac{1}{4\pi\varepsilon_o} \frac{\vec p \cdot |\vec r^{\,}|\hat{r} }{r^3}
Maybe you cope with this!?
Hi,
the electric potential is defined as
\vec E(\vec r) = - \vec \nabla V(\vec r)
in your sketch the electric field vector can arbitrarily be defined as
\vec E(\vec r) = E_0 ~ \vec e_x \qquad \mbox{with} \qquad E_0 = 1000 ~ \mathrm{V}/\mathrm{m}
because only one component of the electric...
i don't know what you mean with ddp.
Despite i suggest that you take a look at the experiment which Richard Tolman conducted in 1916. That would have led you in the right direction.
zerobladex asked for Electric Potential Energy and not for Electrical potential!
V(r) = E - T ~ \ne ~ q_{+} \, \phi(r) \qquad \mbox{with} ~ q_{+} ~ \mbox{as unit POSITIVE charge}
where T is the kinetic energy of the particle and E the total energy.
I don't agree with you! Compare electric potential with electric potential energy. Because two like charges are always repulsing each other the electric potential energy between them is always positive (cause you have to expend work against the force field to get one charge from infinity to any...
In generally the work expended against a force field is positive.
The work expended towards a force field is negative.
With this arbitrary definition we obtain the result you have stated.
So if you have a charge q_1>0 at the origin and a charge q_2<0 at r_2 then the work you must expend on q_2...
yes, the circuits are equivalent! To apply Kirchhoff's voltage law to the circuit in the picture Circuit.JPG it would be helpful to add the ground potential!
Because the capacitor and the resistor are parallel their voltage drop is the same!
I think these circuits are not very instructive!
I assume, that we already know that the plane waves
\vec E(\vec r,t) = \vec E_0 ~ e^{i(\vec k \vec r - \omega t)}
\vec B(\vec r,t) = \vec B_0 ~ e^{i(\vec k \vec r - \omega t)}
are solutions of Maxwell's equation. Let's calculate
\mathrm{rot} ~ \vec E(\vec r,t) = - \frac{\partial...
Hi,
electromagnetic waves have always vector character and a common relation (yield from Maxwell's equations) between the electric field \vec E(\vec r,t) and the magnetic field \vec B(\vec r,t) is
\vec B(\vec r,t) = \frac{1}{\omega} \, \vec k \times \vec E(\vec r,t)
As you stated...
I think i can help you, but with my own labelling ;)
F \sim m \cdot \frac{v^\prime - v}{\delta t}
v^\prime is the speed after a short time interval \delta t related to the instant of time t. So the equation above can be written in a more descriptive way
\delta t \, F \sim m \cdot \Bigl(...
look at Gauss' law for example.
It states, that the flux of \vec E through an arbitrary closed surface, that framed the volume \mathcal{V}, is always
\oint \limits_{\mathcal{S}} \mathrm d\vec S \cdot \vec E(\vec r) \sim Q
That can be proven empirically! For example with a Van de...
Fundamental for your questions is the dispersion relation for electromagnetic waves
\omega^2 = c^2 \cdot \vec k ^2 \qquad \mathrm{or} \qquad \lambda = c \cdot T
and Huygens' principle, that implies, that every point of an existing wave is the origin of a spheric wave with the same...
Although i can't decrypt the formula you have stated, i give it a try
\vec A(\vec r) = \frac{\mu_0 I}{4\pi} \oint \limits_{\mathcal{C}} \mathrm d\vec r^{\, \prime} \, \frac{1}{|\vec r - \vec r^{\, \prime}|}
The curve \mathcal{C} is parameterized through \vec r^{ \, \, \prime}(t) ...
yes the e-field is varying with time! But if the e-field is varying, the b-field must varying with time too.
In the case of time varying fields, maxwell's equations doesn't decouple. Like you see in the expression below, the b-field depends on the time varying e-field and the other way around...
You can easily derive Kirchhoff's rules from the maxwell equations, if you assume that
\frac{\partial \vec B}{\partial t} = 0
then the relevant maxwell equations are
\mathrm{rot} \, \vec E(\vec r) = 0 \qquad \mathrm{;} \qquad \mathrm{div} \, \vec B(\vec r) = 0
Kirchhoff's...