# Tag Info

## New answers tagged potential

0

The line integral would only work for finding a potential difference, which I presume is what you need. Since the line integral needs a path(to perform the integration), hence, it would need a general expression for electric field(in your case, finding the potential requires integrating from $\infty$ to $A$, so you'd need the expression for all points ...

1

The conceptual problem here is that of EMF, $\mathcal{E}$ vs Electric Potential, V. They aren't really the same thing despite being measured in the same units. For instance the EMF is caused by an external agent that isn't the conservative electrostatic field, like say a chemical reaction in a battery or a solar cell. Work is done to cause a charge ...

3

The gravitational field $\mathbf g$ equals, by definition, negative of the gradient of a correspondonding potential $\Phi$; \begin{align} \mathbf g = -\nabla\Phi. \end{align} Therefore, it suffices to produce a gravitational potential $\Phi$ whose value is zero at a point but whose gradient is non-zero at that point. This is straightforward to do. Let a ...

2

What you're calling the "dipolar component of the potential" is not actually that object. For something to be a 'component' of the potential it also needs to be a scalar; in that sense, the sum of those three terms, $$\Delta \phi=\sum_j\frac{\partial \phi}{\partial x_j}(x_j-x_{\text c,j})$$ could indeed be called the 'dipolar component', and it will work in ...

0

The electric field is given by $$E=-\nabla V-\frac{\partial A}{\partial t},$$ where $V$ is the electric potential and $A$ is the vector potential. So you can get the electric field from the potential. Or if you have measurements of a large enough set of points that are close together compared to the scales over which the field varies you might be able to ...

0

If $E< V(x)$ everywhere, and if we assume that the kinetic energy operator $T=\frac{p^{\dagger}p}{2m}$ is a (semi)positive operator, then the TISE implies $$0 ~\leq~ \langle \psi | T | \psi \rangle ~=~ \langle \psi | (E-V) | \psi \rangle~<~ 0,$$ which is impossible. Here $H=T+V$ is the Hamiltonian operator.

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This particle with have an unphysical wave function which blows up (as can be quite easily derived). Therefore, in quantum mechanics, we do not have any particles with $E<V_\text{min}$.

3

OP considers an equations of motion of the form $$\tag{1}\dot{\bf x}~=~{\bf B}({\bf x}),$$ where the vector field ${\bf B}$ is of the form$^1$ $$\tag{2} {\bf B}~=~{\bf \nabla}\times {\bf A}.$$ In other words, ${\bf B}$ is divergence-free $$\tag{3} {\bf \nabla}\cdot {\bf B}~=~0.$$ Eq.(3) is locally eqivalent to eq. (2), cf. Poincare's Lemma. Let ...

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My initial equation was correct by I had neglected to include the modulus signs. Thanks to @jazzwhiz for pointing it out subtly.

1

The multipole coefficients associated with a $1/|r|$ distribution $\rho$ depends on the choice of origin. For example, if you have a point charge and you choose the origin to be at that point charge, then it will have a pure monopole character. However, if you choose the origin to be elsewhere, it will have nonzero expansion coefficients other than the ...

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Introduction $\def\ph{\varphi}\def\vr{\vec{r}}\def\eps{\varepsilon}\def\rmr{{\rm r}}\def\pd{\partial}\def\l{\left}\def\r{\right}\def\ltag#1{\tag{#1}\label{#1}}\def\div{\operatorname{div}}\def\grad{\operatorname{grad}}\def\nR{{\mathbb{R}}}$ You do not need a combination of a surface charge and a volume charge to re-produce a point-charge field outside a ...

0

if the cube is an insulator or a dielectric, how do I find the surface charge density and the volume charge density? Since you are working with an insulator, you can have $uniform$ surface and volume charge densities unlike in the case of a conductor( other than a conducting sphere ). The reason is, excess charge cannot redistribute in an insulator as ...

1

Yes, the uniqueness theorem guarantees it. If you find a solution in which the potential on the surface of the cube looks just like the potential of a point particle, then the potential outside the cube must be identical to that of a point particle since there is only one unique solution satisfying this this boundary condition. (Assuming no external ...

1

Suppose you have a charged cube as shown below, Say it holds a charge of $Q$ Coulomb. Now, for this cube to behave like a point charge, you need to go very $far$ away from it such that at a particular position, your cube appears like a point charge. In this case, the cube $behaves$ just like a point charge with a charge of $Q$ Coulomb. It is clear ...

0

I don't see why there could't be asymmetries for r>c, considering the fact that the charge distribution is not symmetric. So what is the argument for this symmetry? Look at the figure below, In the initial stage, there is an electric field within the conductor( electric field created by the the charge $q$ ) which lasts for an $infinitesimal$ $period$ ...

1

In general, both the positive and negative real exponentials are needed, but there are special cases where only one is needed. If the barrier is "one-sided", that is it looks like a step function, then the region of space where the classical energy would be negative extends to infinity. One of those exponentials will approach zero as distance increases ...

1

Suppose the potential on the outside metal surface is $V_s$. Then the potential outside the sphere is given by the solution of Laplace's equation on that domain. However, it's easy to show that the potential $$V(\mathbf{r})=V_s\frac{c}{|\mathbf{r}|}$$ satisfies Laplace's equation on the domain $|\mathbf{r}|\geq c$. Since solutions to Laplace's equation are ...

3

When you are expanding around a minimum, the derivative is zero by definition. Recall how you find the extrema of a function: the condition $$\frac{df}{dr}=0$$ gives you $r_{\text{min}}$, around which you can then expand.

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