Dipole and test charge

Problem

Dipole and test charge. Two charges $+q=+5\;\text{nC}$ and $-q=-5\;\text{nC}$ are separated by $d=4\;\text{cm}$ along the horizontal segment. A small test charge $q_p=+1\;\text{nC}$ of mass $m=1\cdot 10^{-12}\;\text{kg}$ is placed at point $P$, equidistant $L=10\;\text{cm}$ from the two ends of the dipole as shown in the figure.

(a) Draw the two fields $\vec{E}_+$ and $\vec{E}_-$ at $P$ and their resultant. (b) Calculate the magnitude of the total field $E_P$ at $P$. (c) Estimate the speed that the test charge, released from rest, would reach when the potential has dropped by $\Delta V=1\;\text{mV}$ from its value at $P$.

Solution

(a) Constructing the fields. At $P$ the field $\vec{E}_+$ points away from $+q$ (along the line $+q\to P$); the field $\vec{E}_-$ points towards $-q$ (along $P\to -q$). The two charges are equal in magnitude and equidistant from $P$, so $|\vec{E}_+| = |\vec{E}_-|$. The vertical components cancel, while the horizontal components add: the resultant field $\vec{E}_P$ is horizontal, parallel to the dipole axis, pointing from the $+$ pole towards the $-$ pole.

(b) Magnitude of the total field. Each charge is at distance $L=0{,}10\;\text{m}$ from $P$, so the field of a single charge is $E_1 = k\,\frac{q}{L^2} = (9\cdot 10^9)\,\frac{5\cdot 10^{-9}}{(0{,}10)^2} = 4{,}5\cdot 10^{3}\;\text{V/m}$ Only the horizontal component survives; the geometric factor is $\cos\theta = (d/2)/L = 0{,}02/0{,}10 = 0{,}2$. Adding the two contributions: $E_P = 2\,E_1\cos\theta = 2\,(4{,}5\cdot 10^{3})\,(0{,}2)$ $\ev{E_P \approx 1{,}8\cdot 10^{3}\;\text{V/m}, \quad \vec{E}_P \parallel \text{dipole axis}}$

(c) Speed via conservation of energy. The work done by the field on the test charge converts into kinetic energy: $q_p\,\Delta V = \tfrac12\,m\,v^2 \;\Rightarrow\; v = \sqrt{\frac{2\,q_p\,\Delta V}{m}}$ Substituting $q_p = 1\cdot 10^{-9}\;\text{C}$, $\Delta V = 1\cdot 10^{-3}\;\text{V}$, $m = 1\cdot 10^{-12}\;\text{kg}$: $v = \sqrt{\frac{2\,(1\cdot 10^{-9})\,(1\cdot 10^{-3})}{1\cdot 10^{-12}}} = \sqrt{2}\;\text{m/s}$ $\ev{v \approx 1{,}4\;\text{m/s}}$ The method (force $\vec{F}=q_p\vec{E}$ parallel to the dipole, then energy balance $q_p\Delta V = \tfrac12 mv^2$) is the one required; with the given data the speed reached for a potential drop of just $1\;\text{mV}$ is of order 1 m/s.

Links

Topics: Electric field and potential Concepts: Electric dipole · Electric field Skills: Superposition principle · Conservation of energy