I've just finished reading a section in a book on Coulomb's Law. I'm trying to practice the math a bit and came up with the following:

Suppose you have two charges, -5uC ($q_1$) and 7nC ($q_2$). These charges are 2nm apart and immersed in air. Coulomb's Law is...

$$F = \frac{kq_1q_2}{r^2}$$

Applying my values (using the Coulumb's Consant for air) and evaluating we have:

$$F = \frac{(9*10^9Nm^2/C^2)(5 * 10^{-6}C)(7 * 10^{-9}C)}{(2 * 10^{-7}m)^2}$$
$$F = \frac{(315*10^6N)}{4 * 10^{-14}}$$
$$F = 78.75*10^{20}N$$
$$F = 7.875*10^{21}N$$

These charges should be quite small. While that is the case, the distance between them is also small. That said, it doesn't seem like it should be resulting in $7.875*10^{21}N$.

Supposing I am correct about the large magnitude of the force, and supposing that I have $q_1$ to the left of $q_2$ on an axis ($x$) the vector for $q_1$ would be:

$$v_1 = 7.875*10^{21}N$$

With the vector for $q_2$ being:

$$v_2 = -7.875*10^{21}N$$

Suppose I put a piece of something in between these two charges, assuming the charges to be ions of the same element, do the ions crush whatever is between them, or do they break through it, or? Would that object simply change the force being applied by affecting $k$? Would you have to average the permeability of the air and the material to find $k$?

  • 1
    $\begingroup$ Besides the mistake pointed out by @aquirdturtle, also note that $2$ nm $= 2*10^{-9}$ m, not $2*10^{-7}$ m. The correct answer is thus on the order of $10^13 N$. This is certainly a large force, but a) the distances are really small and b) don't think that because $5$ and $7$ are small numbers $5$ and $7$ C are small charges. The Coulomb is a huge amount of charge compared to everyday experience. For instance, per wikipedia, the typical static charges created by friction (e.g. rubbing a comb on wool) are $\sim 10^{-6}$ C, whereas a typical lightning bolt discharges just $15$ C. $\endgroup$ – ApproximatelyTrue Jun 4 '15 at 7:20

You have miscalculated, an easy mistake to make. In your first line of math, $$F=\frac{315\times 10^{-6}}{4*10^{-14}}$$ not $$F\ne\frac{315\times 10^{6}}{4*10^{-14}}$$ So you are off by 12 orders of magnitude, but this is still a very large force.

Keep in mind what you are suggesting. Electrons have a charge on the order of $10^{-19}$ coulombs, so you are talking about $10^{10}$ and $10^{13}$ electrons. 2 nm is only the distance of 20 or so atoms, so your electron density is absolutely astronomical, which explains your force. No ion would ever have that large a charge. If you were ever able to actually produce such a situation, there is no telling what will happen. The charged objects themselves would probably explode because no known force could hold that many electrons or protons together on an ion.

If you picked more reasonable charges, on the actual scale of ions, the material in between the ions would likely be polarized due to the electric fields in the way that you normally talk about in an E&M course, but you'd have to deal with microscopic defects and the non-continuous nature of the material in-between the ions at this point, so the actual net force calculation would probably be quite difficult.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.