Hot answers tagged electrostatics
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I usually find it easier to use model multipoles that are surface charges on a sphere, rather than point charges on some polyhedron's vertices. These charge densities are given in general by $$\sigma_{lm}(\theta,\phi)=N\cos(m\phi)P_l^m(\cos(\theta)).$$
Thus a monopole is constant, a dipole has $\sigma=\cos(\theta)=z/r$, a quadrupole has the form ...
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When discussing an ideal parallel-plate capacitor, $\sigma$ usually denotes the area charge density of the plate as a whole - that is, the total charge on the plate divided by the area of the plate. There is not one $\sigma$ for the inside surface and a separate $\sigma$ for the outside surface. Or rather, there is, but the $\sigma$ used in textbooks takes ...
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The dipole moment of a system of charges $q_i$ located at positions $\mathbf r_i$ is defined as the vector
$$\mathbf d=\sum_i q_i\mathbf r_i.$$
If you have a single charge $q$ at $\mathbf r=d\hat{\mathbf e}$ then $\mathbf{d}$ has magnitude $qd$ and points along the unit vector $\hat{\mathbf e}$. Usually, however, this is introduced for two charges of equal ...
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Potential energy is a property of the system, not any one object. Thus there should only be one copy of the typical $1/r$ potential energy between two charges (plus an analogous gravitational term if that can't be neglected).
The easiest way to see this is to start from "infinite" separation. Instead of pushing the two charges together, hold one fixed and ...
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There are three different measures of an object's mass: its inertial mass $m_i$ (defined by Newton's second law), its passive gravitational mass $m_p$ (defined by how much force it feels in a gravitational field), and its active gravitational mass $m_a$ (defined by the strength of the gravitational fields it makes). You get qualitatively different ...
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