# Tag Info

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Apparently you are not considering the work done by the electric field, and you are confusing yourself with the "energy storage" concept. After traveling a $\Delta x$ distance into the electric field $\vec{E}$ (avoid using $dx$ because $d$ usually stands for infinitesimal changes), the particle B with charge $q$ will have suffered change in its "horizontal" ...

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Yes you are. If a force is conservative, its work does not depend of any path between any points $A$ and $B$. Since the work integral can depend only on the initial and final points themselves, we define $$W_{A\rightarrow B}=\int_A^B\vec F\cdot d\vec r\equiv U(A)-U(B).$$ Now define the mechanical energy as $E=K+U$ so that $$dE=dK+dU.$$ Suppose there are two ...

1

If voltage is all that you know, then the answer is No. If you know how much charge Q in Coulombs is added, you only have to divide by the charge e on each electron (in Coulombs). Otherwise, if you have a parallel plate capacitor and you know the capacitance C, you can work it out from Q = CV. Your suggestion that the extra electrons 'push more' like a ...

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Try to consider the changes in KE, and how it alternates with EPE and GPE. You should also try and think about what happens to the velocity and acceleration of the mass when it reaches equilibrium, and when it is at maximum amplitude above and below the equilibrium point (consider the energies involved in those 3 specific points). This should give you some ...

1

The analogy would be voltage as the potential, so this would be analogous to height (as in the height of a rock of mass m in gravity field g). Since power (which is also energy) is Voltage x Amps, then Amperes is analogous to mg. If you want to break it down further, I guess mg = Q/sec. I'm not sure if this answers your question, but it at least gets ...

4

In relativity we only use the rest mass, also known as the invariant mass, of an object. In days past the concept of a relativistic mass was used, but this is now strongly deprecated as it has caused endless confusion. For example an obvious question is whether the increase of relativistic mass with speed can cause an object to become a black hole (tl;dr it ...

0

Yes it will do. You can take the potential energy to be zero when the spring is neither compressed or stretched. In special relativity the total invariant mass of the system would then include a contribution from the potential energy / $c^2$. The concept of mass in general relativity is quite subtle, but for weak gravitational fields, the Newtonian limit for ...

0

What is the difference between the potential difference and potential energy of an electron? If I understand your question right, these terms are describing the same thing - one is just in a "per charge" version. Electric potential energy $U_e$ is the potential energy associated with one spot in the circuit. Electric potential or just potential $V$ is ...

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For a discrete system of $N$ charges, the potential energy associated with their configuration is given by \begin{align}U&= \frac12\,\sum_{j=1}^N \, q_j\sum_{k\ne j}\,\frac1{4\pi\epsilon_0}\,\cdot \frac{q_k}{r_{jk}}\\ &= \frac12\,\sum_{j=1}^N \, q_j\,\varphi(\mathbf r_j)\tag 1 \end{align} where $\varphi$ is the scalar potential due to all charges ...

0

You have $${U = \frac{1}{2}\frac{Q^2}{\frac{\epsilon_{0}A}{d}}}$$ $${\frac{\epsilon_{0}A}{d}}=C\equiv\textrm{ the capacitance of the capacitor.}$$ Hence $${U=\frac{1}{2}\frac{Q^2}{C}}$$ Now $\displaystyle{Q=CV}$, where V is the p.d. between the capacitor plates. So, $${U=\frac{1}{2}CV^2}$$ or $${U=\frac{1}{2}\frac{\epsilon_{0}A}{d}V^2}$$ ...

1

Let $X$ be a sphere of radius $R$ with charge $+Q$ on it. The potential of sphere $X$ is $+\frac{kQ}{R}$. Let $Y$ be another sphere of radius $6R$ with charge $+5Q$ on it. The potential of sphere $Y$ is $+\frac{k5Q}{6R}$. The potential of sphere $X$ is larger than that of sphere $Y$ so if the spheres are connected with a wire electrons will flow from ...

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Your approach looks correct. I didn't check your arithmetic, but I suspect that you got the right answer and that the issue here is that your expectations of Interactive Physics are not realistic. Interactive Physics is a simplified educational version of the Computer Aided Engineering program called Adams. This program is intended to produce results ...

0

A non conservative force is generally anything that is affected by friction or air resistance. The amount of energy lost is distance dependent because the further the path traveled, the more an object is affected by these forces. Conservative forces are displacement dependent, meaning they depend only on an initial and final position, and that all the ...

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The simplest example for a non-conservative force is frictional force. For example, to move an object from point A to point B on the table we would require say $10$ J of energy, now to put it back to point A, we would need same $10$ J of energy. Practically nothing observable has changed, but we've performed $20$ J of work, which went to overcome frictional ...

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As Giorgio says, you are overcounting the terms when multiplying by 4. I would like to elaborate a little bit, just to show how this is more than just a math error. It is also a conceptual error: You mention in your reasoning "The total potential energy = the sum of the PE of each of the 4 charges." But this is false. Not only that, but there is no such ...

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This time you're wrong. You are overcounting the terms when multiplying by 4. Potential energy is given for two particles at a time, so you should be multiplying by 2, which is the same as multiplying by 4 and then dividing by 2.

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Potential energy and inertia are related via Einstein's famous equation $E = mc^2$. A compressed spring has more potential energy and therefore more inertial mass compared to an uncompressed one, making it more difficult to accelerate because of its increased inertia.

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Potential energy depends upon the presence of a force, and the actual configuration of objects. So a rock at the top of a cliff has potential energy $U=mgh$, due to $m$, the mass of the rock, $g$, the acceleration due to gravity, and $h$, the height of the cliff. Inertia, as described by Newton's First Law of Motion, is that property of matter which, in ...

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You will get infinity because in addition to $kQ_1Q_2/d$, it also includes the self-energy of the two point charges, which is infinity.

2

A force field is called conservative if its work between any points $A$ and $B$ does not depend on the path. This implies that the work over any closed path (circulation) is zero. This also implies that the force cannot depend explicitly on time. Consider for instance a time decaying force on a straight line. Choose a long closed path. The magnitude of the ...

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Well, contrary to what Lemon said, internal energy is sum of ALL energy of constituting particles related to their DISORDERED motion only. To prove my statement, look out for the derivation of formula for internal energy for an IDEAL gas, and the derivation for value of C(molar heat capacity at constant volume) henceforth. Now coming to your question, since ...

0

To place a charge in the vicinity of an electric field, you should do work against the electrostatic force on the charge. This work done to bring a charge q to an electric field of some other charge configuration from infinity to a distance r, in the field is what we call the potential at the point r. To do a work to move a charge q from a potential V to a ...

1

Relation between Electric field and potential The relationship between electric field $\bf E$ and scalar potential $\varphi$ is given as $$\mathbf E= -\mathbf \nabla\,\varphi$$ where $\mathbf \nabla \equiv \textrm{gradient operator}\;.$ I am unable to understand from this - sign comes. It is worthy to quote from Purcell: The minus sign came in ...

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For your second case, you can change the angular momentum, but remember that you have fixed your total energy. You can't make the planet revolve arbitrarily fast or it will have more energy than allowed. By increasing the angular momentum without adding energy, you are circularizing the orbit. To add, you might take a look at the Specific Orbital Energy ...

1

I guess you mean $$V=\frac{1}{r}$$ instead of $V=1/x$ (if $\partial_y V=\partial_z V=0$ all your integrals diverge). If this is right, then $$V=\frac{1}{r}\Longleftrightarrow \rho\propto \delta(\boldsymbol r)$$ and not $\rho\propto \frac{1}{r^3}$. Anyway, your example $V=1/r$ is highly pathological: it diverges too fast at the origin (the integrals blow up ...

0

Work done is $\vec F \cdot \Delta \vec x$. If $\vec F$ and $\Delta \vec x$ are in the same direction then the work done is positive. If $\vec F$ and $\Delta \vec x$ are in opposite directions then the work done is negative. Consider a spring fixed at one end as a system and an external force $\vec F$ stretching the spring. If the external force \vec ... 0 A single mass attached to a spring has only one degree of freedom. Here you have a diatomic molecule, which in a physical point of view can be visualized as two masses connected by a spring. A single particle has three degrees of freedom. When you attach another particle, that is a constraint, which reduces the overall degree of freedom to two. Now, about ... 0 Potential energy of a body is its capacity to do work by virtue of its position in a conservative force field.So if the body is free to move it will do so in such a way as to reduce its potential energy and the reduction in the potential energy will be equal to the positive work done by the body which could result in the raising of a weight or displacement ... 0 Although there are excellent answers, I think a more "simplistic" answer is required to correct your thinking. If you start with a piece of lead (1 kg) on the floor, grab and lift it 1 meter, it will gain (1 x 9.8 x 1 =) 9.8 J of energy. If you now open your hand (release it), it will fall by "it self" and hit the ground and "loose" the energy it had ... 0 It is just a convention. If we would use the opposite convention, we would get for conservative forces $$\vec F=+\vec\nabla U.$$ You can easily see (think in the one dimensional case) that the particle would move to points of maximal potential energy. This only sounds strange because we are used to the opposite. The mechanical energy principle would retain ... 0 The decrease or increase in potential energy is converted as mechanical work either to increase or decrease it's kinetic energy (we are considering here conservative systems). Suppose you have a system at rest at some height above the ground level, say, a ball placed on the top of a mountain. The work done in order to bring the ball of mass m to a height h ... 0 It's a convention. The real reason is so that we can have: \begin{align} \Delta {\rm KE} &= W_{\rm ext}\\ \Delta {\rm KE} &= W_{\rm con} + W_{\rm noncon}\\ \Delta{\rm KE} &= -\Delta {\rm PE} + W_{\rm noncon}\\ \Delta{\rm KE} + \Delta {\rm PE} &= W_{\rm noncon}\\ \Delta {\rm E} &= W_{\rm noncon} \end{align} Which doesn't work ... 0 It would be better to say that potential energy is the amount of work that a system can do. Say you have a system consisting of two masses - a brick and the Earth. As the brick moves down U decreases, the force pulling the brick and the Earth together acts downwards on the brick and it can do some positive work. On the other hand, if brick moves up, U ... 1V(x)$is a potential energy function for the system of a particle or particles interacting with a set of constraints. These constraints can be thought of as fields which produce a force on the particle(s) of interest. In the infinite square well (ISW), we examine a particle which has no interaction at all until it gets to some impenetrable constaint, i.e., ... 0 The potential$V(x)$used in Schrodinger's equation is used to either model the motion of particles in real systems (in which case the parameters of the potential depend on the particle properties, from Coulomb forces where the property of interest is electric charge and spin or nucleon-nucleon interactions where isospin and spin is important) through to ... 0 The potential energy definitely depends on the properties of the particle. For example, if you were using electric fields to contain the particle, the potential energy would depend on the charge of the particle; if you were using gravity to contain the particle, the potential energy would depend on the mass of the particle. When there is just one particle ... 1 As far as I know, the potential energy depends on the properties of the particle itself (its mass, its charge) No, not really. Think of a classical mass-spring system. The potential energy is $$V(x)=\frac{1}{2}kx^2$$ which is independent of the properties of the mass. In some other cases, such as a charged point particle, the potential energy could be ... 0 Maybe, I found the solution. let's denote with$\theta_k$the rotation around$k$-axis. Suppose that both$\vec{m}$and$\vec{B}$belongs to$xy$plane. That is: $$\vec{m} = \begin{pmatrix} m \cos(\theta_z)\\ m \sin(\theta_z)\\ 0 \end{pmatrix} ~\text{and} ~\vec{B} = \begin{pmatrix} B \\ 0\\ 0 \end{pmatrix}.$$ The torque using the formula$\vec{\tau} = ...

1

Must admit this got me really worried, and I wondered if the problem was maybe in the use of $\bf{\tau} = \bf{r \times F}$ and identifying the force with the (negative) gradient of some potential. Then, I really got worried because it seemed to me that the same problem would occur if you were to make similar calculations with determining the torque on an ...

-1

Potential energy is not energy. If you, for example, place an object of mass m on your outstretched palm and raise it through a distance h, so that it is initially stationary and ends up being stationary, then the total work done on the object by the weight force acting on it is -mgh and the total work done on it by the normal force acting on it due to its ...

1

By work-KE theorem, $$\Delta \text{KE} = \text{Work done by NET force}$$ $$=\text{Work done by} F_1+\text{Work done by} F_2+\cdots$$ Now if a certain force $F_i$ is conservative, you have the choice of defining its corresponding PE so that $$\text{Work done by}F_i=-\Delta \text{PE}_i$$ and MOVE it to the left hand side so that you have ...

0

I want to elaborate on MAFIA's correct and important "potential energy is a property of the whole system". The potential energy is not a property of just one of the involved objects, like the lifted ball. That in our common experience all the potential energy somehow seems to be "attached" to the ball is just a consequence of the very different masses.1 ...

0

The answer is to do with what happens when you stop moving the object upwards. You say that the upwards velocity is 'low enough that it does not leave your hand': the only value of velocity for which that is true is $0$. What this means is that, if at time $t_0$ the object is being moved upwards with velocity $v_0$ and is at $h_0$, and you suddenly stop ...

1

You are not defining your systems clearly enough. If the system is the mass and the Earth then in the situation that you have described the external force doing work on that system increases the potential energy of the mass-Earth system. Now look at the system which is the mass alone. Then there are two forces acting on the mass. The force as described ...

0

If you lift a book, then you are doing work on it. You are actually spending work on raising it's kinetic energy and on lifting it up. $$U=K+W$$ When you reach some height and stop, then all the work done is what determines your height. If you move your hand faster in order to speed up the book to a larger kinetic energy, then you've spent more work on ...

0

Remember that every time we talk about work being done, we must know which force that is doing it as well as how the signs are defined. where $W(begin)$ is the Work done to bring the object from infinity to a certain position in space, and $W(after)$ the contrary, from position to infinity." What force is doing this work? That would be some external ...

0

There are two equivalent ways of defining the potential at a point. The potential at a point is the work done by an external force in bring unit positive charge from the zero of potential (often taken as infinity) to the point. The potential at a point is minus the work done by the electric field in bring unit positive charge from the zero of potential ...

0

In the case of the ball on the table, it is in a state of stable equilibrium, where the table is pushing up against the ball to counteract the gravitational force pulling it down. If the table and ball were to be moved to a planet with much stronger gravity, the gravitational force on the ball could be strong enough to break the table and the ball would move ...

3

This answer is only about Where does the stored energy stay in the object, and why does it only convert into vertical motion and not horizontal motion? because I think your other questions have been well-addressed, but this one has only been answered in highly technical terms that may not have clarified anything for you. Think about what happens if ...

3

The velocity does not double if the acceleration is doubled. The relevant SUVAT equation is: $$v^2 = u^2 + 2as$$ where in this case $u=0$ so we get: $$v = \sqrt{2as}$$ A doubling of acceleration means that the velocity would double if the travel time was kept constant. However in this case it's the travel distance that is held constant. The greater ...

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