# pure dipole vs. physical dipole

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 $kqQ/(x+d)^2$.

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.