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3

In a rechargeable battery, two types of reversible chemical reactions take place: Oxidation reaction: in which a chemical, referred to as the reducing agent ($Re$) is oxidised by donating electrons: $$Re \to Re^{z+} + z e^-$$ Reduction reaction: in which a chemical, referred to as the oxidising agent ($Ox$) is reduced by receiving electrons: $$Ox + z ...


5

Potential energy stored in a body is relative. We have to first choose the potential at a finite point or infinity. In the case given above, we take potential energy to be 0 at the centre of the Earth. So according to the relation, $PE = mgh$ where $h$ is the height from the centre of the Earth. Generally, we take height from the surface of the Earth and ...


3

As indicated in the answer to your previous question, since as $E(=30\:\mathrm{eV})$ is higher than $V_2(=20\:\mathrm{eV})$, that particle is not bound, it's not an eigenstate of the system's Schrödinger equation (it's a scattered state). Its wave function for $x \to +\infty$ would something like: $$\psi=c_1e^{-ikx}+c_2e^{+ikx}$$ ... where both complex parts ...


0

Applying Gauss at a distance $r$ from the centre of the sphere you get $g(r)\;4\pi r^2 = - 4 \pi G M_{\text{enclosed}} \Rightarrow g = - \dfrac{GM_{\text{enclosed}}}{r^2}$ where $g$ is the gravitational field. So all you need to do is to find $M_{\text{enclosed}}$ for each of the regions. What this equation tells you is that you can treat the enclosed mass $...


4

Technically "potential difference" is the difference in electrical potential, i.e. $\Delta V$, not the difference in electrical potential energy, $\Delta U$. Potential difference ($\Delta V$) is also called voltage, in certain contexts. However, many people and sources are sloppy about their terminology, and they will say just "potential" when they really ...


1

Image object $m$ is at some point $a$ and you were to supply a force opposite to the gravitational force caused by $M$. This force is equal to $F_{stop}=-F_g$ so that $m$ hovers completely still at point $a$. M ------------------- a ↑ m Obviously this force can't do any work because the object doesn't move and ...


1

Saying that work done is +ve or -ve is a mathematical convention used for calculating energy transfers. It is +ve when it is done by us on the system, and -ve when it is done by the system on us. Positive work is done in pushing against a force to reach a configuration - eg pushing a car uphill. Negative work is done if a force pulls in the direction we ...


1

The object would experience a net negative force, and be moved a negative displacement. Potential is defined in terms of the work done by an external force. The object has a negative force acting on it due to the gravitational attraction so the external force acting on the object must be in the positive direction to have a net zero force on the object. It ...


1

The work done by an external agency in bringing the body to a point is needed. Note that the process must be quasistatic, otherwise kinetic energy terms will be needed. Now gravity will tend to attract a mass. Thus to keep the process quasistatic, one must oppose this gravitational attraction. In other words, F is directed oposite to the gravitational ...


0

Feynman is using definite small quantities (inches) in place of infinitessimals $\delta x$ etc. Probably he wanted to avoid non-essential mathematical formality, in line with his casual, hand-waving persona. The Principle of Virtual Work requires the structure to undergo infinitessimal displacements (hence "virtual"). He could instead have used units of ...


1

Suppose you move the body down at constant speed, as small as you like, then the net force on the weight will be zero, that is, the force upwards you make will be exactly the same as the weight, $mg$. This upwards force makes negative work, as the displacement is opposite to its direction, and this work is exactly equal to the loss of potential energy, $mgh$....


-1

If the process is quasistatic, no energy is lost as heat, which is obviously impossible in the real case.


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You can compute the integrals $W=\int\vec F\cdot d\vec r=q\int E\cdot d\vec r$ for each charge $q$ you bring from infinity, where $\vec E$ is the electric field generated by the other charges present and then add them all together. On the other hand, since electrostatic fields are conservative the work done is independent of the path, so it is conservative ...


0

Find the potential of the system once the charges have been arranged. Can you relate that to the work done to assemble the system?


1

When you write "energy given out" or "energy gained" you are expressing a choice of sign for the energy transfer that you are specifying, and each of them is different. Notice that in your scenario A loses energy, so it's "energy given out" is positive, but it's "energy gained" would be negative. Employing a sign convention means always stating changes of ...


1

there is an electric field inside the wire, and there is a loss of potential energy, or voltage as they move but this drop in voltage is usually negligible (thought not in some applications) and we only consider that the drop in voltage comes only from the circuits elements o loads. This idealization often fails not with the wires, but within the battery ...


1

You have added the negative sign in front of your integral and then put in the cos(180) as well. Pick one. Since your external force is opposite in direction to the force of the sphere, and you've already put the cos(180) in there, there is no need for the extra negative sign in front.


0

The force exerted on a unit positive charge is $ \dfrac {kq}{r^2} (+\hat r)$ and so an external force $ \dfrac {kq}{r^2} (-\hat r)$ must be applied to move the charge. If the step is $d\vec r$ then the work done by the external force in moving that step is $ \dfrac {kq}{r^2} (-\hat r) \cdot d\vec r = -\dfrac {kq}{r^2} dr$. You have moved from $\vec r$ to $\...



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