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The last part is possibly the most interesting in that you have think about what happens to a dielectric when it is placed in an external electric field, or put another way; how does the movement of charges within the dielectric change the net electric field between the plates?


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You have the right equation. Since you are not given absolute values, you should just reference your answers to the original $Q$. For example, you can say for the first part "The charge will be 0 when the voltage difference is 0. As the voltage difference increases to $\Delta V$, the charge will increase (proportionally) to $Q$". No numbers were needed... ...


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If A would gain kinetic energy, it would move far from B. As A would move more far, Potential Energy of B won't increase as distance had increased proportionally.


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There is no conflict here. Let the two charged particles ($M,Q$) be the system with no external forces acting. Momentum is conserved and so for all time $M_BV = M_A V_{Af} + M_BV_{Bf}$ Energy is also conserved and so for all time $\frac 12 M_B v_B^2 + \dfrac{kQ^2}{R_i} = \frac 12 M_B v_{Bf}^2 + \frac 12 M_A v_{Af}^2 + \dfrac{kQ^2}{R_f}$ The ...


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Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem. Perhaps the confusion comes from the way potential is introduced in ...


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Well there is two answers: 1. It is not infinite. At a point when the black hole has taken in lots of matter, it will throw some out of the black hole to again be able to take in matter.It is like the biggest foodie in the world who can eat a lot but at a point when his tummy is full, he needs to throw up or wait until he can take in more. 2. It is infinite. ...


1

Black holes are in the realm of General Relativity. In GR even the law of conservation of energy is under question when approaching singularities of the GR solution. Potential energy is a concept that comes with conservation of energy. Where the singularity in the black hole solutions is dominating, one cannot talk in terms of energy conservation and ...


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Your question What is the potential energy of a black hole? doesn't make sense because energy is a somewhat tricky concept to deal with in GR. If we treat the black hole as fixed we can study the motion of a test particle falling into it, and we find that there is a quantity analogous to total energy that is constant as the particle falls in. So in ...


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You have got your forces and potentials mixed up. If the force is given by: $$ F(r) = \frac{GMm}{r^n} $$ then the potential is the integral of this: $$ V(r) = -\frac{GMm}{(n-1)r^{n-1}} $$ In the question you link the force has an inverse cubic dependance, $n=3$, while the gravitational force has an inverse square dependance, $n=2$. We get a circular ...


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The reason is mainly in order to be able to write the total energy of the system as $$ E=T+V $$ where $T$ is the kinetic energy of the system. It is far more useful to choose the signs of the terms in the total energy to be $+1$ once and for all, rather than setting the force equal to plus the gradient of the potential. The link between the conventions is ...


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Energy or the value of $V(x)$ negative means it is a bound system. Think of it in this way, if a particle is free and has no kinetic energy and potential energy then it's total energy is zero. If this particle is not free or otherwise is bounded by a negative potential well then it's potential energy is $-V$. You have to give the same amount of energy, in ...


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I think what is missing is the definitions of $F$ and $V$. Consider first the definition of potential energy. The potential energy at a point relative to another point is the work done by a external force (eg the force exerted by you on the mass, $\vec F_{my}$) in taking the mass from the first point to the second point. That force which you exert on the ...


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I think your problem is in how you understand the gradient. Define your z-axis. Lets say it points up. Now $$V(z) = mgz$$. Thus gradient of V(z), that is $$\nabla V(z) = mg \hat z$$. Here, $$\hat z$$ is an unit vector pointing up. So the force will be a vector of length mg, pointing down, because of the minus sign.


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Answer you people have given is right, but it is from mathematical background. Physically, conservative force is a dissipative force. Due to dissipation properties we right as gradient of potential and negative sign comes from its opposite direction of action.


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I also had hard time to picture what Feynman wanted to explain. It feels like my confusion arose from lack of proper definitions of system state, perpetual motion, and reversible machine. The examples he used are also not quite clear of the mechanical apparatus used, not sure if by design he wanted to abstract away that from the reader, but if so, I would ...


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I know this answer is pretty late, but I'm hoping it'll help at least a little. The idea here is, of course, using Green's Theorem: $$\oint \mathbf F \cdot \mathrm{d}\mathbf r = \int_A (\mathbf \nabla \times \mathbf F ) \; \mathrm{d}\mathbf a $$ Here, you're asking about the line integral around a closed path. When using this equation, you end up making ...


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In your case there will be a negative potential on the conductor and the potential on earth is zero. When you ground it there will be a potential difference, so the electrons will move until the potential difference becomes zero ,i.e. electron will move to ground. Then finally the potential on the conductor will also be zero. Since, a neutral metal ball ...


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First there will be a potential on the conductor and the potential on earth is zero. when we ground it there will be a potential difference, so the electrons will move until the potential difference becomes zero. Then finally the potential on the conductor will also be zero.


4

Extremum of potential vs. extremum of action Yes, the action principle is in a special case equivalent to the principle of extremum of potential energy (the maximum of a potential also presents an equilibrium, even though it is an unstable one!). Consider the action principle of a point particle in a potential $V(\vec{x})$, then the Hamilton's action ...


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The values presented in the bar chart aren't really energies, they are changes in energy. Using the frame of reference shown ($y$-axis) the kinetic ($K$) and potential ($U$) energies at point 1 are: $$K_1=0$$ $$U_1=mgy$$ Assuming no friction or air-drag acts on the car, then Conservation of Energy applies, so: $$K+U=\text{constant}=mgy$$ When the car ...


2

Remember that potential energy is defined up to a constant. Here it's defined so that when $r \to \infty$ the potential energy is worth 0, for convenience. So that there's nothing special about a negative potential energy per se. What really matters here is that there's a "well", i.e. a region for $r$ where the potential energy has a minimum and that ...



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