# Tag Info

## New answers tagged potential

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You're trying to understand how a point charge is represented in the charge density $\rho(r)$. The concept you need is the Dirac delta function $\delta(r)$, which can describe the density of a finite amount of stuff packed into an infinitesimal point: $$\delta(r) = \left\{ ^{\infty \text{ if } r=0 }_{0 \text{ if } r\ne0} \right\}$$ $$\int_{-\infty}^{+\... 0 The equation for multiple point charges can be calculated from the equation for a single point charge. The equation for a single point charge is (Where V is voltage, Q is a point charge, r is radius/distance, k is the constant of 1 over 4 pi epsilon and epsilon is the permittivity of free space.) Then, the equation for multiple point charges is ... 0 This turns out to have a really boring answer. We can find the two circular orbits by finding the maxima and minima of the effective potential, and we get:$$ r = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - \frac{12M^2}{L^2}}\right) \tag{1} $$where the + gives the outer stable orbit and the - gives the inner unstable orbit. Note that both orbits exist only ... 1 The values of E_c and E_v in the band diagram depend on the point of reference. So yes they can have negative values if you chose your reference that way. Keep in mind that their difference E_g stays constant nonetheless. Electrons are fermions and therefore governed by Fermi-Dirac statistics. That means that they have to comply with the Pauli ... 2 Short (but cryptic) answer: complex numbers arise in quantum mechanics because we would like find solutions to the differential equation$$\frac{\partial}{\partial x}f(x) = cf(x)$$which don't blow up as x\to \pm\infty. Long answer: Fundamentally, the shift from classical mechanics to quantum mechanics is replacing functions (observables) and numbers (... 0 Imaginary numbers are very useful for calculations, but as their name suggests - they are imaginary, not real. So if you measure something in the real world, you can only expect to get real numbers. This is true for both electrodynamics and QM, expectation values of quantum mechanics observables (i.e. measurements) will always turn out real. Imaginary ... 2 Gravitational field is a vector field and is determined by negative gradient of the gravitational potential.$$\vec g=-\vec\nabla \phi$$Frome equation above, it is obvious that |\vec g|=|-\vec\nabla \phi| (magnitude of \vec g is equal to magnitude of -\vec \nabla \phi) and we know that |-\vec\nabla \phi| is a non-negative quantity. You have made a ... -2 Yes, any constant should be ignored. -1 As the potential is equal, there will be equal amount of work done by the electrons to move from infinity to any point on the surface. If the surface has different potential, then the electron will be accelerated, then it will automatically become a non equipotential surface. Thats why it should be normal. 4 The equation will become$$ - \frac{\hbar}{2m} \bigg( \frac{\partial^2}{\partial x^2} + \frac{ \partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} \bigg) \psi(x,y,z) = (E-V_0) \psi(x,y,z) $$And the solutions are the same:$$ \psi_{n_x, n_y, n_z} (x,y,z) = C \sin( \frac{n_x \pi x}{L}) \sin (\frac{n_y \pi y}{L}) \sin(\frac{n_z \pi z}{L}). $$And ... 1 Charges at rest move when a force is applied on them and this is due to Newton's laws. Now to apply a force, we need a field, like electric/gravitational field. Each field acts upon certain measurable properties of a system, like gravitational on mass, electric on charge etc. Now potential is just a fancy name of height in electromagnetism. I hope you're ... 0 If the charges on the plates of the capacitor are +Q and -Q then the PD between the plates is V = Q/C where C is capacitance. (This is the definition of capacitance.) If we add the same amount of charge Q'>Q to each plate, this has no effect on the PD between the plates, because it will increase the absolute potential of each plate by the same amount. The ... 1 Let us say plate A has a charge q1 and plate B, which faces plate A has a charge q2. By making use of the fact that the net field in the bulk of a conductor in static conditions is zero, and that the net field near the outer surface of a conductor equals [local surface charge density/€0], you can prove the following: Charge on the outer surfaces of A and B ... 0 Yes there is. Notice that the expression$$ \frac{d}{dt}\frac{\partial{T}}{\partial\dot{q}_\alpha}-\frac{\partial{T}}{\partial{q}_\alpha}+\frac{\partial{V}}{\partial{q}_\alpha}=0 $$is very close to the Euler-Lagrange equations (ELeq)$$ \frac{d}{dt}\frac{\partial{L}}{\partial\dot{q}_\alpha}-\frac{\partial{L}}{\partial{q}_\alpha}=0  except that the ...

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If you have a proton (mass=$m_p$) and a neutron (mass=$m_n$) and allow them to join via the strong nuclear force the mass of the $^2_1H$ nucleus $m_{np}$ is less than the sum of the masses of the individual particles ($m_{np} \lt m_p + m_n$). During this joining together the proton-neutron system loses potential energy and that energy is called the binding ...

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Mass certainly will change. In fact we have two connected bodies - brick and earth. The Hamiltonian (full energy operator) of a connected particle system is defined as: $H = H(brick) + H(earth) + V$, where $V$ is energy of interaction (potential energy in our case). Since potential energy will increase, mass inevitably increases due to the theory of ...

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No, mass does not increase when you lift an object. Potential energy increases, but that's because of the $\Delta h$ in $U_{\mathrm{grav}} = mg\Delta h$, not the $m$. Mass is, to put it a bit imprecisely, the amount of matter in an object. When you lift something, there doesn't become more of it. (Note that I'm ignoring relativistic effects.)

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