Homework Statement
What is the correct interpretation of
< \frac{\partial {A}}{\partial t} >, where A is an operator?Homework Equations
for a wave function \phi and operator A,
<A> = \int_{V}\phi^{*}(A\phi)dVThe Attempt at a Solution
I thought it could mean
< \frac{\partial {A}}{\partial t}...
I think I know Gauss' Law for electric fields, but I still do not see how it can be used to find the boundary conditions.
I know that Gauss's Law states that the total charge enclosed by any closed surface equals the sum of components of the electric field perpendicular to the surface. I...
thank you for the quick reply
the potential vanishes as r approaches infinity, so I guess that means C4 =0
but there was no boundary condition given for the edge of the sphere. that is why I'm having trouble... :/
Homework Statement
A total charge q is uniformly distributed throughout a sphere of radius a.
Find the electric potential in the region where r1<a and r2>a.
The potential is defined anywhere inside the sphere.
Homework Equations
letting ρ = volume charge density and ε = permittivity...
yes, you're right, thank you :smile:
I agree with the restriction you are suggesting, and I think it can be mathematically stated as follows:
<-4,0,6>\cdot\vec{r} = 0 (r is perpendicular to <-4,0,6>)
-4r_{x}+6r_{z}= 0
2r_{x}-3r_{z}= 0
OR<-4,0,6>\cdot\vec{F} = 0 (F is perpendicular to...
even if there were a way to do the cross products backwards, you still wouldn't get what you want because you are looking for two unknowns r and F with only one equation
r x F = <-4, 0, 6>
if you provide either r or F, then you might be able to find a way to do the cross product "backwards"...
yes, the inductive step does require k ≥2 to be valid, and if that's the case, the base case should be k =2 not k=1; and even if we use the base case k=2, the theorem still cannot be proven
base case k =1:
the proof is flawed because it fails to carry out the inductive step
(1)S(1) is valid but...
i think, in general, it is necessary.
"domino" analogy: even if the first domino falls, if it does not cause the second domino to fall, then one cannot guarantee that the rest of the dominoes would fall.
yes, we can start from n =1. This completes the first step of the proof. However, a proof by induction requires 2 statements, which you outlined in your first post, to be true.
Both statements MUST be true. however, in this theorem, statement (2) does not hold for some k (k=1), that is P_{k}...
Thank you! :smile:
no, the implication P_1 \rightarrow P_{2} is not true (if there exists a set of 1 blonde girl whose member does not have blue eyes).
Is the following proof convincing?
Suppose, that P_1 \rightarrow P_{2} if there exists a set of 1 blonde girl whose member G1 does not have...
:smile:
so i was reading an old thread that mentioned the following problem from Apostol's Calculus I
I'm not sure if I my solution is correct, so I am posting it here for opinions:
The inductive step is correct, i.e. if it can be shown that the theorem holds for n=3, then the theorem is...
yes, I've done that
∂Ex(x) / ∂x = ε0ρ(x) for 0<x<d
and
∂Ex(x) / ∂x = 0 for 0>x and x>d
Ex (x)= ε0∫ρ(x)dx (from 0 to x) + Ex(0) for 0<x<d
and
Ex (x)= Ex(0) for x<0
and
Ex (x)= Ex(d) for x>d
my problem is how to arrive at the conclusion that
Ex(0) = Ex(d) ?
Thank you for the reply.
We haven't covered the Poisson equation yet. We've covered Maxwell's equations though, both integral and differential forms and the boundary conditions. So, I was thinking if there's another way without using that Poisson equation...
But thanks, i'll check that. :smile: