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  1. J

    Physics Problem: Fnet=ma or Fnet= 0?

    My suggestion to you is to remember that according to Newton's Laws of Force, a body in motion stays in motion and an object at rest stays at rest, unless acted upon by an external force. F = ma, and F = 0 are the mathematical representations of this statement. F = 0 implies the system's...
  2. J

    Partial differentiation problem, multiple variables (chain rule?)

    You should try the steps I previously mentioned. Then reflect on why it works..
  3. J

    Partial differentiation problem, multiple variables (chain rule?)

    Try saying z = x2 + 2\frac{x^2}{cos^2θ}sin2θ = x2(1 + 2tan2θ) Now with this for z you can perform \left(\frac{\partial z}{\partial \theta}\right)_{x} quite easily. Hint* Remember, that after you perform the derivation to look for anywhere you can make a substitution to remove 'x'.
  4. J

    Evaluate the surface integral

    Now I would like some clarification...because I don't see how the projection on the surface could be zero when there is no symmetry argument for the point at the top of the hemispherical bowl.
  5. J

    Evaluate the surface integral

    My cylinder example was derived from a line charge along an axis. I hadn't read the problem until he attempted the integration. A disk is still applicable however, as to close the surface on the plane intersecting the normal would be the same.
  6. J

    Evaluate the surface integral

    It seems to me he did account for the surface? I picked a disk and told him to use the divergence theorem because I assumed it an easier route to take.
  7. J

    Evaluate the surface integral

    My fault, I didn't read the problem. You're right, as that is the volume of a single hemisphere.
  8. J

    Evaluate the surface integral

    You're off by a factor of 2. You're integral for Phi should be 0< Phi < pi.
  9. J

    Evaluate the surface integral

    No, that's right. With the disk, the volume would be that of a cylinder. So you're divergence = 1 implies "a" is growing linearly in a single direction. So the RHS of the equation would be something like (using the disk analogy), z*2pi*R, which is the volume of a cylinder. Both sides hold.
  10. J

    Evaluate the surface integral

    Picturing it in 2D with a disk may be the best way to learn this, then apply the concepts to 3D. Imagine a flat disk in the XY plane. dS would be a vector along the z-axis whose magnitude is equal to a small area of the disk. If integrating this dS, to just S (the surface), one would have a...
  11. J

    General solution to PDE

    I think he just made a typo.
  12. J

    General solution to PDE

    I found a mistake here, \frac{∂f}{∂y} = \frac{∂f}{∂u}\frac{∂u}{∂y}+\frac{∂f}{∂v}\frac{∂v}{∂y}, not what it says in the quote.
  13. J

    General solution to PDE

    Since you factored out a \frac{∂}{∂x}, this implies the right hand side is constant with respect to a change in x. If this were an ODE, it's implied the value is constant, but because you have a function of two variables, it is implied to be a function that is constant with respect to a change...
  14. J

    General solution to PDE

    First look at the equation: \frac{\partial^2 f}{\partial x^2}+4\frac{\partial^2 f}{\partial x \partial y}+\frac{\partial f}{\partial x}=0 We can factor a \frac{∂}{∂x} out and write it as such: \frac{∂}{∂x}(\frac{∂f}{∂x}+4\frac{∂f}{∂y}+f(x,y)) = 0. Now what this implies is...
  15. J

    Integration by parts with orthogonality relation

    You're answer is wrong because \int_{0}^{a} sin\frac{n\pi x}{a}sin\frac{m\pi x}{a}dx = \begin{cases} \frac{x}{2} &\mbox{if } n = m; \\ 0 & \mbox{otherwise.} \end{cases} The \frac{a}{2} is after evaluation of the integral from 0 to a. So your \intvdu is actually \int\frac{x}{2}dx.
  16. J

    Unsure about Inverse Laplace Heaviside Function question

    Yes Chet, it is, but not a 5...it's a 5/6 i believe. I was just trying not to give him f(t-a) :p So: \frac{5}{6}uc(t)sin(6(t-8))
  17. J

    Unsure about Inverse Laplace Heaviside Function question

    It's because your understanding of the Heaviside Step Function is lacking definition. The fact that you have \inte-8s(\frac{5}{s^{2}+36})ds is the same as saying you have a square wave with the function F(s) convoluted with it. When you integrate this, it's like integrating a Dirac-Delta...
  18. J

    Force on an electric dipole

    If the problem was not electrostatic and void of bound charges, then the equation would not simplify. You'd have a TON of terms.
  19. J

    Force on an electric dipole

    The answer is in my previous post...the math is done component wise once the simplifications have been made.
  20. J

    Electrostatic fields- What defines an electrostatic field?

    Well the divergence would remain unchanged, but yes, a curl would develop. This curl is a time-varying magnetic field. Charges in motion produce time-varying magnetic fields!
  21. J

    Electrostatic fields- What defines an electrostatic field?

    Correct, now go one step further and think about what the divergence implies about the physical charge in space, and what you would expect to happen when you move the charge. Compare what you think to Maxwell's Equations (Gauss' Law and Faraday's Law)
  22. J

    Electrostatic fields- What defines an electrostatic field?

    Just think of what a static field means. Think of what it implies about the direction of the field lines. If I have a stationary charge (static), what would I expect its divergence to look like? What about its curl?
  23. J

    Force on an electric dipole

    If \vec{F} = -\vec{∇}U, and U = -\vec{p}\cdot\vec{E}, then \vec{F} = \vec{∇}(\vec{p}\cdot\vec{E}). Now from vector calculus: \vec{∇}(\vec{p}\cdot\vec{E}) = (\vec{p}\cdot\vec{∇})\vec{E} + (\vec{E}\cdot\vec{∇})\vec{p} + \vec{p}x(\vec{∇}x \vec{E}) + \vec{E}x(\vec{∇}x \vec{p}) Now since the dipole...
  24. J

    Objects falling into a black hole

    We do see them glow brightly at the event horizon, in the form of an accretion disk, as well as at the poles, where large quantities of x-rays can be emitted. Observing the actual black hole is not possible with visible light, but measuring the orbital speeds of visible matter and calculating...
  25. J

    Infinite potential

    Yes region (I) has a 0 wave function. Region (II) requires a solution to the Schr. equation, but it should be similar to that.
  26. J

    The continuity equation in electromagnetism

    Not really a different definition, but yes it is a different viewpoint. You seem to be grasping the relationship, and understanding how both \frac{dQ}{dt} is zero and current constant. E&M can be very confusing when you first develop these relationships, but the more you work with it and...
  27. J

    The continuity equation in electromagnetism

    Exactly. Remember, E&M is relativistic. The trouble is when you first start E&M, they treat the electric and magnetic fields as separate entities, when they really are not. Edit: To elaborate, from your viewpoint of the wire, there is a constant current flowing across it. From the viewpoint...
  28. J

    The continuity equation in electromagnetism

    You just seem to be confusing a single point charge in motion I = \frac{dQ}{dt} with a current density \vec{I} = λ\vec{v}. A single point charge in motion is meaningless to examine unless we care about the fields \vec{B} and \vec{E} associated with it. Every example you have described is...
  29. J

    The continuity equation in electromagnetism

    You're correct. Not quite. ∫S \vec{J}\cdotd\vec{S} = \intS limA→0\frac{\vec{I}}{A}\cdotd\vec{A} As A trends toward zero in the limit, it cancels the differential dA. This means as you approach the area of a point (point charge), the surface disappears. So again, we are looking at a...
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