Recent content by saunderson

  1. S

    Examples of isothermal processes in everyday life

    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...
  2. S

    Justification of a common calculation

    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 :rolleyes: Now my thoughts about this are about to come to maturity... Any other suggestions are appreciated as well =)
  3. S

    Justification of a common calculation

    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...
  4. S

    Magnetic Field Around Finite Conductor: Questions & Answers

    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.
  5. S

    Magnetic Field Around Finite Conductor: Questions & Answers

    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...
  6. S

    Seeking Exercises to Accompany Feynman's Lectures on Physics

    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...
  7. S

    Here we go again Newtons and Watts

    Physically you do no work because you are physically a rigid body! At least at the macroscopic level.
  8. S

    Here we go again Newtons and Watts

    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...
  9. S

    Here we go again Newtons and Watts

    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 FMaybe it is helpful for you to...
  10. S

    Here we go again Newtons and Watts

    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...
  11. S

    Derivation of energy stored in a capacitor

    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...
  12. S

    Derivation of energy stored in a capacitor

    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) ?
  13. S

    Evaluating Electric Dipole Potential: qd cos \theta?

    Absolutely right! I hope that is a sufficient answer to your question!?
  14. S

    Evaluating Electric Dipole Potential: qd cos \theta?

    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} =...
  15. S

    Evaluating Electric Dipole Potential: qd cos \theta?

    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!?
Back
Top