# Electric Field Lines

This post aims to clarify the following passage (Serway 727).

We choose the number of field lines starting from any positively charged object to be $C'q$ and the number of lines ending on any negatively charged object to be $C'\left|q\right|,$ where $C'$ is any arbitrary proportionality constant. Once $C'$ is chosen, the number of lines is fixed. For example, if object 1 has charge $Q_1$ and object 2 has charge $Q_2,$ then the ratio of number of lines is $\frac{N_2}{N_1}=\frac{Q_2}{Q_1}.$

First, let us try at a better explanation, rather than a translation, of this passage. Since charge is quantized (Serway 710), it follows that given two charges of magnitudes $Q_1$ and $Q_2,$ we must have $Q_1=k_1e$ and $Q_2=k_2e,$ for some integers $k_1$ and $k_2.$ It follows that $Q_1=\frac{k_1}{k_2}Q_2,$ or $\frac{Q_1}{Q_2}=\frac{k_1}{k_2}.$ Hence, $Q_1$ and $Q_2$ are commensurable since their ratio is that of two integers. In other words, both charges can always be expressed as integral multiples of a common charge $q,$ the least of which is $e.$ That is, a charge $q$ exists such that $Q_1=k_1q$ and $Q_2=k_2q,$ for some integers $k_1$ and $k_2.$ We arbitrarily choose the number $N_i$ of field lines we want for $Q_i\ (i=1\ \mathrm{or}\ 2)$ so that $N_i=nk_i$ ($n$ a positive integer). That is, the number of field lines for $Q_i$ can be any arbitrary integral multiple of $k_i.$ The number of lines attached to the other charge $Q_j$ is then given by solving $\frac{N_1}{N_2}=\frac{k_1}{k_2}$ for $N_j.$ The result is $N_j=nk_j.$ For example, if $\frac{Q_1}{Q_2}=\frac{3}{4},$ then the number of fields lines must satisfy $N_1=3n$ and $N_2=4n$ for any positive integer $n.$ In summary, we have the following result:

If $\frac{Q_1}{Q_2}=\frac{k_1}{k_2}$ for some integers $k_1$ and $k_2,$ then for any positive integer $n,$ if $N_i=nk_i$ is the number of field lines attached to $Q_i,$ then $N_j=nk_j$ must be the number of field lines attached to $Q_j.$Fig. 2. “Better” explanation of Fig. 1.

The following attempt at translating the passage should make it clear that the above explanation is "much better". From the preceding discussion, $Q_1$ and $Q_2$ are commensurable so can be expressed as integral multiples of some common charge $q$ as follows: \begin{align} Q_1&=n_1q\\ Q_1&=n_2q \end{align} This is the $q$ being referred to in the passage. We then have: \begin{align} C'q &= N_1 =\mathrm{number\ of\ lines}\\ C' &= \frac{N_1}{q} =\mathrm{number\ of\ lines\ per\ unit\ charge} \end{align}

Using the example given in the last paragraph of (Serway 727), let $Q_1=-q$ and $Q_2=2q$ where $q>0.$ Choosing the same value as the author we have $N_1=C'q=8,$ or $C'=8/q.$ That is, we have decided to have 8 lines for each charge $q.$ With $C'$ determined this way $N_2=C'\left(2q\right) =\left(\frac{8}{q}\right)\left(2q\right) =16.$ That is, 8 lines attached to $Q_1$ and 16 lines attached to $Q_2.$ Notice that by the first explanation, we could have simply found the ratio of field lines $\frac{Q_2}{Q_1}=\frac{2q}{q}=\frac{2}{1},$ and then chosen $N_2=16,\ \left(n=8,\ k_2=2\right),$ giving us $N_1=8,\ \left(nk_1=\left(8\right)\left(1\right)=8\right).$ This is verified by the proportion $16:8::2:1$ which is true. In other words, the proportionality constant given by $\frac{Q_i}{Q_j}$ seems to be more useful than $C'$ in determining how many lines to have. Evidence that the excerpt was simply poorly worded is suggested by the author's choice of the symbol $C'$ as the proportionality constant. The prime serves no purpose in the excerpt, he could have simply used $C.$

## Works Cited

Serway, Raymond A. and Robert J. Beichner. Physics for Scientists and Engineers. 5th Edition. Brooks/Cole, 2000. Print.