# Electrostatic force between two grains of sand with unbalanced charge placed 30 meters apart

Quote from The Feynman Lectures on Physics Volume I:

To give an idea of how much stronger electricity is than gravitation, consider two grains of sand, a millimeter across, thirty meters apart. If the force between them were not balanced, if everything attracted everything else instead of likes repelling, so that there were no cancellation, how much force would there be? There would be a force of three million tons between the two!

I am trying to work out the computation but I am unable to get the number three million tons in my result.

Here is my computation with some reasonable assumptions I have made.

The molar mass of silicon dioxide is $$60.08 g/mol$$. Let us assume its density is $$2.6 g/cm^3$$. Let us assume that the grain of sand that is a millimeter across has a volume of $$1 mm^3$$. Therefore a grain of sand has these made silicon dioxide molecules:

$$2.6 \frac{g}{cm^3} \times \frac{6.022 \times 10^23}{60.08g} = 2.6 \frac{g}{1000 mm^3} \times \frac{6.022 \times 10^{23}}{60.08g} = 2.6 \times 10^{19}/mm^3$$

A molecule of silicon dioxide has 30 protons and 30 electrons (8 from each of the two oxygen atoms and 14 from the silicon atoms).

To satsify the following assumption in the book,

if everything attracted everything else instead of likes repelling

we will only compute the attraction between the protons in the first grain of sand and the electrons in the second grain of sand. We will consider the electrons in the first grain fo sand and protons in the second grain of sand to be absent for this purpose.

Since there are $$2.6 * 10^{19}$$ molecules of silicon dioxide in a grain of sand that has $$1 mm^3$$ volume, the total charge due to electrons in the first grain of sand is:

$$-2.6 \times 10^{19} \times 30 \times 1.602 \times 10^{-19} C.$$

Therefore, the total force of attraction between the electrons in the first grain of sand and the protons in the second grain of sand is with a distance of 30 metres between them.

\begin{align*} F &= \frac{8.99 \times 10^9 \frac{Nm^2}{C^2} \times -(2.6 \times 10^{19} \times 30 \times 1.602 \times 10^{-19} C)^2} {(30 m)^2} \\ &= -155966530449.6 N. \end{align*}

Since 1 ton weight = 9806.65 N,

$$155966530449.6 N = \frac{155966530449.6}{9806.65} \text{ ton weight} = 15904159.97 \text{ ton weight}$$

The number 15904159.97 is approximately 15 million. I don't see it getting anywhere close to 3 million as written in the book.

Can you help in fixing my calculation so that it agrees with the force of three million tons as written in the book?

• Your calculation seems to be correct. I get 16.6 million tonnes. If all particles attract, as Feynman assumes, this must be multiplied by 4, so the result is 66.4 million tonnes. – my2cts Oct 14 '18 at 10:36
• The volume of a sphere with diameter 1mm is about 0.52mm^3 not 1mm^3, which gets you a bit nearer to Feynman's number. – alephzero Oct 14 '18 at 11:39
• Feynman is making an order of magnitude estimate. Your answer and his actually agree to within an order of magnitude. – user4552 Jul 2 '19 at 12:14

Your 2.6 value is still in $$\frac{g}{mm^3}$$, whereas your other constants and distances use meters only. You will have to fix this. The volume for a half mm radius will be $$vol = 5.2 \cdot 10^{-10} m^3$$, the density, assuming quartz, will be $$den = 2.6 \cdot 10^6 \frac{g}{m^3}$$. So from here you have for the total number of particles in the volume, $$particleNum = den \cdot vol \cdot (60.08)^{-1}\cdot 6.02\cdot 10^{23}\cdot 30 = 4.06\cdot 10^{20}$$. The total charge will be, $$charge = particleNum \cdot 1.6\cdot 10^{-19} = 64.96C$$. From here the force in newtons will be, $$F = \frac{1}{4\cdot \pi \cdot 8.85\cdot 10^{-12}}\cdot \frac{(charge)^2}{(30)^2} = 4.2\cdot 10^{10}$$, which translates to 4 million tons of force.
• $10^10$ is 10 billion, not a million. I don't think there's an error in the OP's use of units. In your answer, you're writing all the numbers without units, which is a bad idea. – user4552 Jul 2 '19 at 12:13