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2

If you increase the resistance of the bulb by 10 ohm, the total resistance becomes 240 + 10 = 250 ohm. The voltage across this combined resistance remains the same(120 v). So, you just need to plug this new resistance value in the formula and that will give you the power being consumed by the combined resistance which gives us 57.6 Watt. As we can see that ...

0

The Kronig-Penney model is a simplified version of atoms on a lattice which is made by replacing the potential by an array of rectangular barriers. We actually do take the same method used to solve the finite potential barrier step, i.e write down wave-like solutions of the Schroedinger equation and solve for the boundary conditions. The only new things here ...

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Current do flow in neutral objects, it's just that net current is zero. Now, consider that A is +ve and B is -ve. Then when you connect them, free electrons of B are attracted by +ve charged A, which you said as both like 0 potential. While truth is that there is no tendency towards 0 potential but there is just electrostatic attractions giving current.

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There is an alternative version of Ohm's law $\vec J = \sigma \vec E$ where $\vec J$ is the current density (current per unit area), $\sigma$ is the conductivity of the material and $\vec E$ the electric field strength. If $\vec E$ did become zero then no current would flow. If for a fixed $\vec E$ the conductivity increases (resistivity/resistance ...

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When we bring a test charge let us say (+q) to a certain point, we exert some force. Our exerted force has to be equal or greater to the force exerted by the electric field in order to overcome it. Thus the forces being opposite and equal cancel each other and hence the direction is undetermined. Since it is a compulsion for vector to have both magnitude and ...

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It can be easily seen using the Heisenberg picture. Take a Hermitian operator $A$ that commutes with the Hamiltonian $H$. Remember that the eigenvalues of $A$ are observables. Then $A$ also commutes with the time evolution operator $U(t) = e^{-i H t}$. $[A,U(t)] = 0 \quad \rightarrow AU(t) = U(t)A \rightarrow A = U(t)AU(t)^\dagger$ So the operator $A$ ...

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If you have an eigenstate of an operator $A$: $$A|Ψ(t=0)>=a|Ψ(t=0)>$$ Then apply $A$ to the time evolution of the wavefunction: $$A|Ψ(t)>=Ae^{-iHt}|Ψ(t=0)>=e^{-iHt}A|Ψ(t=0)>=e^{-iHt}A|Ψ(t=0)>=a|Ψ(t)>$$ Note that the second to last equality is only true if $[A, H]=0$. So the eigenstate of $A$ remains an eigenstate of $A$! This applies ...

2

A conserved quantity is one that commutes with the Hamiltonian for the simple reason that $[A,H] = 0$ implies $$\frac{\mathrm{d}}{\mathrm{d}t} A = 0$$ in the Heisenberg picture. Another way to see that commuting with the Hamiltonian means conservation is to consider that the time evolution operator $U(t) = \exp(-\mathrm{i}Ht)$ is just the exponential of ...

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Yes. If you have two positively charged plates of different charge magnitudes and fire a stream of electrons between them, the electrons will deflect toward the "more positive" plate. Though really the electron stream isn't exactly necessary, since the two plates will repel each other indicating a potential difference right off the bat.

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If by magnitude difference you mean Plate A has a charge of +1 and plate B has a charge of +9, then there will be a potential difference. If by magnitude difference you mean Plate A has a charge of +1 and plate B has a charge of +1, then there won't be a potential difference.

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It all depends by what you mean by the word "flow". Let the charged body which is assumed to be a conductor produce an E-field. In a conductor which has mobile charge carriers then the charges can be made to flow within a conducting body which has no net charge. If you subject an uncharged conducting body to an external E-field then the mobile charge ...

3

It sounds like you're asking "If two conductive materials are brought in contact, and one of them is electrically neutral, and one is positively charged, which direction will charge flow?" If that is indeed your question, then the answer is that negative charges (electrons) will flow from the neutral object to the positive object until they are at the same ...

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Potential at center due to +ve sphere is not correct. What you had found is when cavity is at center. However potential due to $-\rho$ is correct. First consider no cavity Potential at center of sphere due to uniformly charged complete sphere $V = 3kq/2a$ Now, potential due to positive charged sphere $cavity$ at center. $$V_{1}= \frac{4\rho π ... 1 You have got your forces and potentials mixed up. If the force is given by:$$ F(r) = \frac{GMm}{r^n} $$then the potential is the integral of this:$$ V(r) = -\frac{GMm}{(n-1)r^{n-1}} $$In the question you link the force has an inverse cubic dependance, n=3, while the gravitational force has an inverse square dependance, n=2. We get a circular ... 3 First of all, don't think of multipole moments as separate things that have their own individual meaning. Instead, think of them as parts of one thing. Once we have all the parts written down, we can start naming and organizing each one to determine its contribution to the whole. Now, for your question Is there a physical interpretation for multipole ... 0 I'm basically at a loss how to simplify the integral. I am not sure this needs to be done. In spherically symmetric cases, the volume element, dV is quite simple to deal with. In your case, you have:$$ \begin{align} W & = \frac{1}{2} \int dV \ \rho\left(r\right) \ \Phi\left(r\right) \\ & = \frac{4 \pi \ C}{2} \int dr \ r^{2} \left( r^{-2} ...

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No, the circuit is not complete. If it were complete Capacitors would discharge and 1st plate would have 0 potential. Next, potential of capacitor and plate are different. Capacitor has potential difference b/w two plates. Since both are connected in series net Capacitance = 10/3 uF . $$Q = CV = 10/3 *100 = 1000/3 uC$$ which is charge on both capacitors. ...

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I think that the arrangement looks something like this?

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In general you're right; if we know $V(0,0, z)$ we can only get its $z$-derivative. But the author is implicitly assuming the electric field is directed along the $z$ axis, because of the rotational symmetry of the problem. Therefore, the $z$ component of the gradient is all we care about.

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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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$\Delta$ is the Laplacian, the sum of second derivatives. It is applied to a function $f$ of the spatial coordinates which you typically want to solve for. $\rho$ is some given function of the spatial coordinates. As mentioned by Martin, Poisson's equation has a multitude of applications throughout most of physics. A cornerstone example is electrostatics ...

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Because it's an excellent approximation. The gravitational potential from a spherical mass $M$ of radius $R$ is, to second order \begin{align} U(r) & = -\frac{GM}{r} \\ & = U(R)+\frac{dU}{dr}(r-R)+\frac12\frac{d^2U}{dr^2}(r-R)^2 +O\left(\left(\frac{r-R}{R}\right)^3\right) \\ & = -\frac{GM}{R}+\frac{GM}{R^2}(r-R)-\frac{GM}{R^3}(r-R)^2 ...

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