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2

The first one is correct. The problem is that in general the force acting on the particle will no longer be conservative in the moving reference frame and we can no longer associate the potential energy $U$ to it. To see why this is so, realize that a force $\vec{F}$ is conservative if and only if: $$\oint_C\vec{F}\cdot d\vec{r}=0$$ to any closed curve $C$, ...


1

Just for the sake of having an answer: In $\phi \nabla ^2 \phi$, consider the term $\phi \frac{\partial ^2 \phi}{\partial x^2}$. By the product rule, this is equal to $\frac{\partial}{\partial x} \left(\phi \frac{\partial \phi}{\partial x}\right) - \left(\frac{\partial \phi}{\partial x}\right)^2$. Combining all three terms, we get $\phi \nabla^2 \phi = ...


1

There are several (equivalent) ways to look at this. One is to say that for any conservative force $\mathbf{F}$, one can define the potential energy Ep as an associated potential field such as $\mathbf{F}=-\frac{\partial Ep}{\partial r}$, or maybe more formally $\mathbf{F}=-\nabla(Ep)$. That's no more than a definition of the potential energy. ...


-2

electric field strength is $$E=\frac Fq=\frac Vd$$ with $V$=voltage, $d$=distance between charged plates \begin{align} \frac Fq&=\frac Vd \\ Fd&=qV \end{align} but $Fd$=energy $$\therefore {\rm energy}=qV$$


0

Here is one take on how to understand the relation between force and potential energy, which I think is the closest modern version of how it would have been seen originally. Let's take as the conditions for a force to be conservative $$\nabla \times \mathbf{F} = 0$$ and $\mathbf{F}$ is a function of position only (this leaves out the magnetic force on a ...


1

We can start at the relationship: $W=-\Delta U$, which is work done by a conservative force. The math A (conservative) force $F$ will do this work on an object when doing a displacement $\Delta x$, and $W=F \Delta x$. In the general case, the force might be different at different points as the object is moved (the force of gravity is not constant along ...


0

If $E < V\left(-\infty\right)$, $E<V\left(+\infty\right)$, and $E > V_{min}$ (necessary for $\Psi$ to be normalizable), then it is a bound state, and the spectrum will be discrete: $$ \Psi\left(x,t\right) = \sum_n c_n \Psi_n\left(x,t\right). $$ Otherwise -- if $E > V\left(-\infty\right)$ or $E > V\left(+\infty\right)$ -- it is a scattering ...


2

This definition of bound and scattering states is not quite correct, although it holds for many potentials. There are counterexamples to this fact that have roots in a paper by von Neumann and Wigner. One is the spherical potential $$V(r)=32\sin r \frac{\sin r-(1+r)\cos r}{\left(2+2r-\sin 2r\right)^2}$$ It is not hard to check that $V(r)$ is a bounded ...


2

To escape Earth, you'd need to be alive for (see this) $$ \frac{1}{365 E} \left(\frac{GMm}{R}\right) \ \mbox{years} \approx 1 \ \mbox{year}, $$ where $E \approx 2000 \times 1000 \times 4 \ J = 8 \times 10^6 \ J$ is the average daily energy intake (couldn't find a good source for a worldwide average), $G \approx 7 \times 10^{-11} \ m^3 kg^{-1} s^{-2}$ is the ...


3

Some quick scribbling suggests that caloric intake for a $70\;{\rm kg}$ human over the course of $50\;{\rm yr}$ at $2500\;{\rm Cal}$ per day is enough to accelerate that person to almost $75,000\;{\rm m}/{\rm s}$. Escape velocity from Earth is about $11,000\;{\rm m}/{\rm s}$, so even with a reasonable hit for inefficiency (about $2\%$ energy efficiency is ...


2

First, one of the implications of the electrostatic potential satisfying Laplace's equation is that the extremes are at the boundaries. If the potential of the surface of the sphere is zero and the potential at infinity is zero, the only solution for the potential outside the sphere is the trivial solution, i.e., the potential is zero everywhere outside the ...


0

On the one hand, we let our reference voltage be at infinity. On the other, we say the sphere is grounded. "Grounded" could mean two things: The potential of the sphere is defined to be the reference. This contradicts the previous definition of the reference to be at infinity. The sphere is connected by an ideal conductor to an infinite source of charge ...


0

But your height on the hill is Rearth + H, where Rearth is the radius of the Earth and, so would U = mg(Rearth + H) (never mind the integral needed for now)? But wait, the Earth is in the potential well of the Sun so shouldn't U = mg(sun)(Rorbit - H)? Luckily, you don't need to know an absolute potential which would require knowing the distances, masses, ...


2

It's known that the general solution of the Laplace equation $\nabla^2\Phi = 0$ in spherical coordinates with azimutal symmetry is given by: $$\Phi(r,\theta)=\sum_{l=0}^\infty\left(A_l r^l + B_lr^{-(l+1)}\right)P_l(\cos \theta),$$ where $A_l$ and $B_l$ are given by boundary conditions. We also have that the function $\dfrac{1}{|\mathbf{x}-\mathbf{x'}|}$, ...


1

We all know that $E=k_E + V(x)$; $E$ = total energy, $k_E$ = kinetic energy, $V(x)$ = potential One reason why you may be confused by this is that the equation $$ E = \frac{1}{2}m\dot{x}^2 + V(x) $$ comes from classical mechanics. When we are solving the Schroedinger equation $$ H\Phi = E\Phi $$ with Hamiltonian operator $H$, the meaning of the ...


1

Since $k_E \propto \hat{p}^2$ and $\hat{p}$ is Hermitian you may see that this makes $k_E$ positive semidefinite, that is all of its eigenvalues are larger or equal to 0. In other words when you measure this operator you will always get results which are larger or greater than zero. This "contradiction" is resolved by the fact that the potential is a ...


2

"the gravitational potential energy should DECREASE as it is converted into kinetic energy, but I guess I'm not really sure why it's negative at position 3." To try and explain this I'll use an analogy. Imagine for a second you have a cat, and you throw the cat up into the air. If you choose to use the ground as a reference point, then the cat starts off ...


0

Furthermore, if PE decreases as r gets larger, then why does a pen gain PE as you lift it further away from a table and why does the PE decrease/convert to KE when you drop the pen? These are two entirely different contexts. In the first context, the potential energy of the system of two objects is considered. In this context, the magnitude of the ...


1

First of all, you wrote the equation for $U$ wrong; it should be $r$ instead of $r^2$ in the denominator. However, that typo isn't what the problem is. The problem is that you've overlooked the word "magnitude" in the question. If a negative number is changed to be a different negative number that's closer to zero, then the magnitude of the number has ...


3

What @lemon said is right. While $\bf F = -\nabla \phi\left({\bf r}\right)$ suggests that $\bf F$ depends on $\bf r$, it ultimately depends on time through ${\bf r} = {\bf r}\left(t\right)$. That is, ${\bf F} = -\nabla \phi\left({\bf r}\left(t\right)\right)$. So, there is no inconsistency in writing Newton's second law as (a dot represents a time derivative) ...


1

"Calculate the trajectory" just means calculate $x(t)$, given the potential energy and the initial conditions.


3

I too was confused by this difference between gravity and electromagnetism. Hopefully the following clears things up. The gravitational potential a distance $r$ from a mass $M$ is $$ \phi_g=-\frac{GM}{r}, $$ the gravitational field is $$ {\bf g} = - \nabla \phi_g, $$ and the gravitational potential energy (of two masses $M$ and $m$ separated by a distance ...


1

Well, the electric field $\vec E$ is different from the force field $\vec F$ a test charge will feel. That difference is exactly the charge of the test particle. That force field is given by the gradient of a function, too $$ q \vec E = \vec F = - \frac{\mathrm d}{\mathrm d r} W$$ where I use the letter $W$ in order not to have confusing notation. The ...


4

I think you are reading a lot into what is a minor distinction. Strictly speaking I suppose the gravitational potential is the energy per unit mass, i.e. $m=1$ in your first equation, while the gravitational potential energy is the potential times the mass. In practice no-one I know has ever bothered to make the distinction because it's usually obvious what ...


0

From Griffith section 2.4.4 comments on Electrostatic Energy, you can get your answer. If you consider point charges, then actually, this integral is related with self-energy which is infinite at usual, to make finite we often introduce cutoff radius $\delta$. (In particle physics, we often use bare and renormalized terminology, renormalization is a some ...


3

Hint: You're far enough from the ground that the formula $mgh$, which is an approximation, doesn't really apply anymore. What is the true formula for gravitational potential energy?



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