# Tag Info

## New answers tagged dipole

0

If the total charge of the system is zero, the dipole moment does not depend on distance. much same like: if total momentum of a system is zero, the angular momentum does not depend on the origin of reference. Dipole moment is the intrinsic property of a system (subtract total charge to zero first); Angular momentum is the intrinsic property of a system ...

3

The electric dipole moment is defined as $$p = \int r \; dq$$ In the case of a pair of charges for which both charges are of the same magnitude, the choice of the origin turns out to be irrelevant: $$p = \mathbf{r_1} q - \mathbf{r_2} q = q(\mathbf{r_1} - \mathbf{r_2}) = q\mathbf{d}$$ where $\mathbf{d}$ is the distance between the charges. However, when ...

1

If you take a permanent magnet, and place a sheet of paper over it. Now sprinkle iron filings on it, and you pretty much get this diagram. This has been the mainstay of field theory since Faraday's time. A test charge at rest will begin to move in the direction of the field line. Since there is nowhere that it can rest where there is more than one ...

0

At any point the electric field is the vector sum of the fields from the two charges. So while the fields from $A$ and $B$ are indeed in opposite directions at your point $p$ you just add them (well, subtract their magnitudes since they're in opposite directions) and this gives you the net field. I wouldn't take the field lines too seriously. They are not ...

1

Dipole $\def\vp{{\vec p}}\def\ve{{\vec e}}\def\l{\left}\def\r{\right}\def\vr{{\vec r}}\def\ph{\varphi}\def\eps{\varepsilon}\def\grad{\operatorname{grad}}\def\vE{{\vec E}}$ $\vp:=\ve Ql$ constant $l\rightarrow 0$, $Q\rightarrow\infty$. \begin{align} \ph(\vr,\vr') &= \lim_{l\rightarrow0}\frac{Ql\ve\cdot\ve}{4\pi\eps_0 l}\l(\frac{1}{|\vr-\vr'-\ve\frac ...

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