দ্বিমেরুর জন্য বিভব

Suppose that you have two opposite point charges $+q$ and $-q$ separated by a distance $d.$ The dipole moment $\vec{p}$ of such a dipole is the product $q\vec{d}$, where $\vec{d}$ is the vector pointing from the negative to the positive charge. Let $p = \left|\vec{p}\right|.$ Let's find the electric potential at every point on the same line as the dipole.

Let the positive charge be at a position $x = +\frac{d}{2}$ and the negative charge be at a position $x = -\frac{d}{2}.$ Then, the electric potential at a point $P = (x, 0)$ is: $$V = \frac{kq}{\left|x - \frac{d}{2}\right|} - \frac{kq}{\left|x + \frac{d}{2}\right|}$$ Breaking into cases and simplifying gives us: $$x\lt -\frac{d}{2}\rightarrow V = -\frac{4kqd}{4x^2-d^2}$$ $$-\frac{d}{2}\lt x\lt +\frac{d}{2}\rightarrow V = \frac{8kqx}{d^2-4x^2}$$ $$x\gt +\frac{d}{2}\rightarrow V = \frac{4kqd}{4x^2-d^2}$$ An interesting thing is to note what happens when $x$ grows very large compared to $d$--either the point moves very far away from the dipole, or the charges become very close to each other, or both. In this case: $$|x| >> d, x\lt 0 --> V\approx -\frac{kqd}{x^2} = -\frac{kp}{x^2}$$ $$|x| >> d, x\gt 0 --> V\approx \frac{kqd}{x^2} = \frac{kp}{x^2}$$ Two interesting things present themselves. The first is that the dipole potential is directly proportional to its dipole moment; this is one reason that the dipole moment is an important property of a dipole. The second is that the electric potential is inversely proportional to the distance squared, as opposed to the distance (as it is for a point charge). This shows that the dipole potential decreases faster than the point charge potential as one moves away from the charge(s), which makes sense, since from far away, the difference in position between the positive charge and negative charge becomes increasingly insignificant, and so the charges seem to cancel each other out.