phyzguy,
A sketch would include four currents, two from the field (electrons move against the field, holes with the field) and two due to random thermal motions (electrons from P to N, holes from N to P). In equilibrium these effects cancel.
Consider a PN junction doped with say phosphorous on the N side, and Boron on the P side. Initially, there is an opportunity for the electrons just below the N conduction band to drop to the lower available energy states just above the P valence band. This leaves the N side positively charged...
To answer my own question, we are comparing the frequency measured by the ground observer -- who is at rest relative to the medium air -- with that measured by an observer moving with the siren and at rest relative to the air. Since they are both at rest relative to the air, they will measure...
Consider the situation where an observer at rest on the ground measures the frequency of a siren which is moving away from the observer at speed ##v_{Ex}##. Let ##v_w## be the speed of the sound wave. Let ##\lambda_0##, ##f_0##, ##\lambda_D##, and ##f_D## be the wavelengths and frequencies...
If what I have is a valid transformation, then I'm confused about what is calculating. Given the coordinates in S, does it give the coordinates in S(bar)?
I noticed that of course. But, isn't it true that the matrix gives S(bar) in terms of S? I don't see how the logic could be wrong. We give S(bar) in terms of S' and S' in terms of S, thus S(bar) in terms of S. I agree the formula is not a boost, but isn't it still correct?
Summary: The problem is to generalize the Lorentz transformation to two dimensions.
Relevant Equations
Lorentz Transformation along the positive x-axis:
$$ \begin{pmatrix}
\bar{x^0} \\
\bar{x^1} \\
\bar{x^2} \\
\bar{x^3} \\
\end{pmatrix} =
\begin{pmatrix}
\gamma & -\gamma \beta & 0 & 0 \\...
The formula for the integral is correct and produces the correct angular momentum. Yet, using the other coordinate system, I'm left with an integral which is divergent according to Mathematica. I mean the system with the origin as the midpoint between the charges. You can try writing down the...
Excellent answer, however in your example of the point mass, you are finding the angular momentum relative to 1. Where the particle itself is by placing it at the origin. 2. Some other location. In my example, I found the angular momentum relative to 1. The location of the electric charge. 2...
Relevant Equations:
Angular momentum density stored in an electromagnetic field: $$\vec{l}_{em} = \epsilon_0[\vec{r} \times (\vec{E} \times \vec{B})]$$
Electric field of an electric charge: $$\frac{q_e}{4\pi\epsilon_0}\frac{r - r'}{|r - r'|^3}$$
Magnetic field of a magnetic charge...
This question is motivated by Problem 7.12 in Griffiths Electrodynamics book. I have not included it in the homework section, because I have already solved it correctly. However, I question whether my solution which agrees with the solution's manual is correct.
Relevant Equations:
$$\Phi =...
Thanks for taking a look at it. The dimensions are correct. I'm using relativistic units, so |a| has units of 1/s. I'll email the professor and see if he is in agreement with our answers.
Homework Statement
Suppose a spaceship starts from rest from a space station floating in deep space and accelerates at a rate of |a| relative to the space station for 1.0 Ms. It then decelerates for the same amount of time at the same constant rate |a| to arrive at rest at another space station...
OK, I see your point. The friction argument shouldn't matter anyways, because the result still holds in a friction free environment. Of course what I meant was the net forces acting on the person, so friction from the ground plus the two Tensions all sum to 0. But, I like the way you are...
On second thought, I don't think I should ignore the frictional force with the ground. The frictional force with the ground is acting say downward, and for static equilibrium, the upward forces from the chains have to cancel, and I think I get the right answer this way. Is that how you thought...
Of course I've done that. If we ignore the frictional forces with the ground as the book usually does in a situation like this, and cancel the gravitational and normal forces, there are two forces acting on the person along the two chains.
I created the thread. I can't see how I'm missing a factor of 2. You should like my thread, it's all in easily to follow algebraic symbols with the calculation at the end.
Homework Statement
A person would like to pull a car out of a ditch. This person ties one end of a chain to the car's bumper and wraps the other end around a tree so that the chain is taut. The person then pulls on the chain perpendicular to its length. If the distance between the car and tree...
Thanks for helping me with that problem. I'm doing self study just for the fun of it. I don't want to start another thread, but I'm confused about another problem on the same topic, and was wondering if you have any insight about it. The problem reads: A person would like to pull a car out of a...
Let d_b be the distance from the floor to the bucket along the slope which is given to be 2.5m. d_s be the distance from the floor to the support along the slope of the ladder. I found this value to be about 1.3462m by using similar triangles. Let theta be the angle between the left leg and...
It's measured up the ladder slope.
I considered the division of weight between the two halves, but that would still be off. I don't think you're supposed to make that assumption. I believe the weight rests and the top of the triangle so that they both receive the 20 kg.
Of course I have. The diagram is an isosceles triangle. That's how I calculated the distance from ground to the support along the ladder and the other calculations.
Homework Statement
Suppose we have a folding ladder, so that when its legs are spread it makes an isosceles triangle. Suppose it has a support that runs parallel to the ground. There is a bucket of paint that rests on top of the ladder, and the mass of the ladder is negligible.
Data: The legs...
If the Iron ball and water have different initial temperatures (and the book certainly didn't say they have the same initial temperature) then delta temperature iron = T_f - T_i(Iron) which is not equal to T_f - T_i(water). They would certainly share the same final temperature after an...
No, the problem does not indicate that. I already thought about that before.
I just redid the problem where I assume that the "Iron" ball does not undergo any temperature change. I almost get the exact same answer. I think we're supposed to assume that the "Iron" ball does not undergo any...
I don't understand your comment. For example, we worked a similar problem where a lead ball crashes into the ground, and we assumed the ground doesn't have any temperature change at all. Here, we are assuming the lead ball and water have the same temperature change, and I don't see why.
Homework Statement
Suppose we drop a lead ball of mass M into water of mass m from a height h and allow everything to settle down. What is the temperature change of the water? Assume that the container is well insulated.
Homework Equations
Potential Energy = mgh
dU = mcdT
where dU is the...