# Unit conversion between dynes and newtons

I am learning Electricity and Magnetism by myself. Can someone explain me the math behind the equation below? (Quoted from Edward Purcell's Electricity and Magnetism (Cambridge University Press, Massachussetts, 2012), 3rd ed.)

$$9\cdot 10^{14} \:\mathrm{dyne} = \frac{(N\:\mathrm{esu} )^2}{(100\:\mathrm{cm} )^2} \implies N^2 = 9\cdot 10^{18} \implies N = 3\cdot 10^{9}$$

I don't understand why is it multiplied by $$10^4$$ to obtain $$N^2 = 9 \times 10^{18}$$.

• There is no dyne to newton conversion here. Just that in your equation, you are multipying both sides by 10^4 cm^2, Just simple algebra. May 3 '19 at 18:36
• Yes however I am curious for the reason to multiply by 10^4. Is it some kind of esu conversion factor? May 3 '19 at 18:39
• to solve the equation and obtain the value of N, you need to multiply it by the denominator on both sides. It is not a conversion of units. May 3 '19 at 18:57
• Note that we use MathJax to typeset mathematics; you can find a good tutorial here. May 3 '19 at 18:58

## 2 Answers

F= $$(k • Q1 • Q2)/ r^2$$ = $$(Coulomb)^2/(metre)^2$$ = $$(N esu)^2/ (100 centimetre)^2$$ $$(Coulomb)^2/(metre)^2$$=9*10^(14) dyne (given) therefore, $$(N esu)^2/ (100 centimetre)^2$$=9*10^14 dyne or, $$(N esu)^2$$= (9*10^14)(100^2) dyne centimetre or, N= $$3*10^9$$ dyne centimetre (esu)

Basically, we found out the relation between 1 Coloumb and 1 electrostatical unit, which is, 1C= $$3*10^9$$ esu.

This passage is not converting between dynes and newtons.

Instead, it is finding the charge $$Q$$ that will produce a force $$F = 9\cdot 10^{14} \:\mathrm{dyne}$$ when two point charges of charge $$Q$$ are set a distance $$r = 100\:\mathrm{cm}$$ apart, and Purcell is writing the value of the charge as $$Q = N \:\mathrm{esu}$$, i.e. $$N$$ is the numerical value of $$Q$$ expressed in the unit $$\rm esu$$ (and has nothing to do with newton's).

In these manipulations, the factor of $$10^4$$ is the square of the $$100$$ in the denominator, i.e. the square of the numerical value of $$r$$ when it is expressed in the unit $$\rm cm$$.

To be frank, the unit manipulations in that passage ─ and, particularly, the fact that Purcell drops the units ─ is more confusing than anything else, basically due to the age of the text. In more modern, and less confusing, notation, I would write \begin{align} F & = k_C\frac{Q^2}{r^2} \\ \implies Q & = \sqrt{F r^2 /k_C} = \sqrt{ \frac{9 \times 10^{14} \mathrm{dyn} \times100^2 \: \mathrm{cm}^2}{1 \:\mathrm{dyn} \: \mathrm{esu}^{-2} \: \mathrm{cm}^{-2}}} = \sqrt{9\times 10^{18}} \:\mathrm{esu} \\ & = 3\times 10^{9} \:\mathrm{esu} , \end{align} with an explicit Coulomb constant.

To be frank, though, the fact that the Coulomb constant is actually dimensionless (i.e. that in ESU units the esu is dimensionally equivalent as $$1\:\mathrm{esu} = 1\: \sqrt{\mathrm{dyn}} \: \mathrm{cm}$$, with a fractional unit exponent and all) is so much of a nightmare that it is enough of a reason to drop the ESU entirely and switch to SI units. (For more on the subject, see my answer to Why are electrical units (specifically, electrical current) considered a base unit?.) If Purcell were still alive, I suspect that (like David Jackson) he would have dropped their "gentlemen's pact" not to give in to SI and made the switch as well (see the introduction to the newer editions of Jackson for more details). This is not to bash Purcell as a resource to learn the concepts (it is a fantastic book), but it does show its age in its handling of the numerical calculations.

• Thanks it really helps! May 3 '19 at 21:22
• @pythonheadache If this answers your question, you can click the checkmark to the left to mark it as accepted. May 3 '19 at 21:36