Irodov 3.2: One per Cent Is Enormous – Amit & Deepali Sinha Foundation

Problem Irodov 3.2

Two copper spheres of mass $1$ g each are $1$ m apart. Suppose that in each sphere the total charge of the electrons differed in magnitude from the total charge of the nuclei by one per cent. How large would the force between the spheres be? Copper has atomic number $29$ and molar mass $63.5$ g/mol.

Given

$m = 1.0$ g each
$r = 1.0$ m
imbalance of one per cent
copper, $Z = 29$ , $M = 63.5$ g/mol (table values)

Find

the force

F

between the spheres

Solution

Step 1The charge of the nuclei in one sphere

Copper is named without its atomic data; the table values are the atomic number $Z = 29$ and the molar mass $M = 63.5$ g/mol. A sphere of mass $m$ holds $m/M$ moles, so the number of atoms is

N = \frac{m}{M}\,N_A = \frac{1.0}{63.5}\times 6.02\times10^{23} = 9.48\times10^{21}

Each nucleus carries the charge $Ze$ , so the nuclei of one sphere together carry

Q_n = ZeN = 29\times(1.60\times10^{-19})\times(9.48\times10^{21}) = 4.40\times10^{4}\ \text{C} \tag{1}

Step 2The unbalanced charge and the force

The electrons’ charge differs from (1) by the fraction $\eta = 0.01$ , so each sphere carries a net charge

q = \eta\,Q_n = \frac{\eta\,Z e\,N_A\,m}{M} = 0.01\times 4.40\times10^{4} = 440\ \text{C} \tag{2}

The two spheres of mass m, centres a distance r apart, each with the same net charge (drawn positive) and pushed apart by the same force F. — The two spheres of mass $m$ , centres a distance $r$ apart, each with the same net charge (drawn positive) and pushed apart by the same force $\vect{F}$ .

A sphere much smaller than $r$ acts as a point charge at its centre, and Coulomb’s law with (2) gives the force:

F = \frac{1}{4\pi\varepsilon_0}\,\frac{q^2}{r^2} = (9.0\times10^{9})\,\frac{(440)^2}{(1.0)^2} = 1.7\times10^{15}\ \text{N}

F = \frac{1}{4\pi\varepsilon_0}\left(\frac{\eta Z e N_A m}{M r}\right)^2 \approx 1.7\times10^{15}\ \text{N}

(3)

Check

The charge of the nuclei, (1), follows independently from the Faraday constant $F_a = eN_A = 96\,500$ C/mol, the charge of one mole of elementary charges: a sphere holds $m/M$ moles of atoms, each with $Z$ such charges, so

Q_n = Z F_a\,\frac{m}{M} = 29\times 96\,500\times\frac{1.0}{63.5} = 4.41\times10^{4}\ \text{C}

in agreement with (1). Units: $q^2/(\varepsilon_0 r^2)$ is in $\text{C}^2/(\text{F/m}\cdot\text{m}^2) = \text{C}^2/(\text{F\,m}) = \text{N}$ , since a farad is a coulomb per volt and a volt a joule per coulomb.

Remark

The force grows as the square of the imbalance, so even an imbalance of one part in $10^{9}$ would scale the force down by the factor $(10^{-7})^2$ , to about 17 N, nearly two thousand times the spheres’ weight. Ordinary matter is neutral to far better than that, which is why the electric forces between everyday objects, overwhelming inside atoms, almost never show on the large scale.