So, let's assume that I have an electric dipole that comprises of charges $+q$ and $-q$ with separation of d placed in a non uniform electric field induced by charge $+Q$ which is a distance $x$ away from $+q$. If I am told to calculate the net force exerted on the dipole by this non uniform field, would my calculations be different if the dipole is pure or physical?
How I would intuitively go about solving this problem, regardless of the type of dipole in question, is by summing up the two vectors representing the electric forces exerted by $+Q$ on $+q$ and $-q$ respectively, such that the magnitude of first force would be $kqQ/(x^2)$ and the magnitude of the second one will be $kqQ/(x+d)^2$.
Can someone please advise?
Best Answer
This depends on the relative sizes of the two length scales ─ the internal separation $d$ of the dipole, versus the length scales at which the external field changes.
The general result for a point dipole is presented in most textbooks, but the basic idea is simple enough. Start off with a finite dipole: you have two charges, $-q$ at $\mathbf r_0$ and $+q$ at $\mathbf r_0+d\hat{\mathbf n}$, and an electric field $\mathbf E(\mathbf r)$ acting on the two, so the total force is simply \begin{align} \mathbf F & = q\mathbf E(\mathbf r_0+d\hat{\mathbf n})-q\mathbf E(\mathbf r_0). \end{align} Now, if the external field is uniform, then the two will cancel out, so you need an inhomogeneous field. If the variation is gentle enough, then a linear variation should be sufficient, so we can take the first-order Taylor series of the electric field at the positive charge: \begin{align} \mathbf F & = q\mathbf E(\mathbf r_0+d\hat{\mathbf n})-q\mathbf E(\mathbf r_0) \\ & = q\left[\mathbf E(\mathbf r_0)+d\hat{\mathbf n}\cdot\nabla\mathbf E(\mathbf r_0)+\mathcal O\left(d^2\frac{\partial^2E}{\partial r^2}\right)\right]-q\mathbf E(\mathbf r_0) \\ & = qd\hat{\mathbf n}\cdot\nabla\mathbf E(\mathbf r_0)+\mathcal O\left(d^2\frac{\partial^2E}{\partial r^2}\right) \\ & = \mathbf p\cdot\nabla\mathbf E(\mathbf r_0)+\mathcal O\left(d^2\frac{\partial^2E}{\partial r^2}\right), \end{align} i.e. if the second-order terms can be neglected, then the only aspect of the charge distribution that matters is the dipole moment, i.e., you can just treat it as a point dipole and forget about it. And, for a point dipole, as $d\to0$ while the length scale of the higher-order changes in the external electric field (signified by $\frac{\partial^2E}{\partial r^2}$) stays put, then the first-order term will become exact.
This is what a point dipole really is, in practice: a finite dipole that is so small compared to any other relevant length scale in your configuration that you might as well take it to be infinitely small. (Well, OK, not quite: you can indeed find some configurations of charge that are 'purely dipolar' and which retain that character even at finite size.)
The final question that remains, then, is what happens when the length scale of the external field's variation is comparable to, or smaller than, the dipole's internal separation. In that case, there are no applicable approximations, and you just have to stick to the initial exact result, \begin{align} \mathbf F & = q\mathbf E(\mathbf r_+)-q\mathbf E(\mathbf r_-). \end{align}