Question

Because the charges on the electron and proton have the same absolute value, atoms are electrically neutral. Suppose that this is not precisely true, and the absolute value of the charge of the electron is less than the charge of the proton by 0.00100%. Estimate what the net charge of this textbook would be under these circumstances. Make any assumptions you feel are justified, but state clearly what they are. (Hint: Most of the atoms in this textbook have equal numbers of electrons, protons, and neutrons.) What would be the magnitude of the electric force between two textbooks placed 5.0 m apart? Would this force be attractive or repulsive?

Short answer

The magnitude of the electric force between the two textbooks is

The force is repulsive in nature.

Step-by-step solution

Determination of the net charge of books.

Number of protons= Number on electrons.

Assume the mass of the book is 1.0 kg

The given value of mass of proton is equal to the mass of a neutron with a value of

$1.67\times {10}^{-27}\mathrm{kg}$

So, the number of protons is,

localid="1663833953627" $\frac{1}{2\times 1.67\times {10}^{-27}\mathrm{kg}}=3\times {10}^{26}{\mathrm{kg}}^{-1}$

Thus, the net charge difference is,

$$\begin{gathered} ∆\mathrm{q}=3\mathrm{x}{10}^{26}\mathrm{c}\mathrm{x}00001\mathrm{x}1.6\mathrm{x}{10}^{-19}\mathrm{C} \\ =480\mathrm{C} \end{gathered}$$

Determination of the magnitude of the electric force between the textbooks.

According to Coulomb’s Law,

$\begin{array}{l}\mathrm{F}=\frac{\mathrm{k}\times {\mathrm{Q}}_1\times {\mathrm{Q}}_2}{{\mathrm{d}}^2}\\ =\frac{\mathrm{k}\times ∆{\mathrm{q}}^2}{{\mathrm{d}}^2}\\ =\frac{\mathrm{k}\times {480}^2}{5^2}\\ =8.3\times {10}^{13}\mathrm{N}\end{array}$

Thus, the force is repulsive in nature.