# Tag Info

5

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 ...

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 ...

3

The center of mass continues to move at the same velocity. You can see this more easily if you make a setup with just two equal masses, and look at the problem in the center of mass frame. If you push them apart, they will move by equal and opposite amounts; when you bring them back together, they will again move by equal and opposite amounts. This means ...

2

Yes, in the sense that the charge carriers in the superconductor will experience a force when there's a changing flux.

2

Physicists have a slightly different (and in my opinion incorrect) way of deriving nodal circuit diagrams than engineers, which would make his statement correct because the circuit diagram would imply that the wires form loops of a finite area, but this way of thinking breaks the usefulness of circuit diagrams in fundamentally undesirable ways. If you ...

2

OK, in this video you've kindly provided me, Lewin essentially talks about this circuit: The easy rule of thumb that's common to all electrical engineers is to say: a current $I$ goes through this loop, causing a voltage drop $R I$ across the resistor, a voltage drop $\int_0^t dt~I(t)/C$ across the capacitor (assuming it is uncharged at $t=0$), and a ...

1

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.

1

I think that the arrangement looks something like this?

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 ...

1

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 ...

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