Question

Two engineering students, John with a mass of $\mathbf{90}\mathbf{}\mathbf{\text{\hspace{0.17em}kg}}$and Mary with a mass of$\mathbf{45}\mathbf{}\mathbf{\text{\hspace{0.17em}kg}}$, are$\mathbf{30}\mathbf{\text{m}}$apart. Suppose each has a$\mathbf{0}\mathbf{.01}\mathit{\%}$imbalance in the amount of positive and negative charge, one student being positive and the other negative.Find the order of magnitude of the electrostatic force of attraction between them by replacing each student with a sphere of water having the same mass as the student.

Short answer

The electrostatic force of attraction is of the order of${10}^{18}\text{\hspace{0.17em}N}$ .

Step-by-step solution

Given

• John’s mass is 90 kg and Mary’s mass is 45 kg, and the distance between them is 30 m.

• There is a difference of 0.01% in the value of the positive and negative charges

Quantization of charge

The principle states that the smallest possible charge that can exist freely is the elementary charge $\mathbf{(}\mathbf{e}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{\times }{\mathbf{10}}^{-19}\mathbf{C}\mathbf{)}$. So, all the charges that exist in nature will always be multiple of the elementary charge. This means that the charge in nature is quantized.

Calculatethe order of magnitude of the electrostatic force of attraction

The net charge $\left(q\right)$on John, whose mass $\left(m\right)$is given as-

$q=\frac{0.01}{100}\left(\frac{m{N}_{A}Ze}{M}\right)$

Here,i$Z$s the number of electron-proton pairs present in a molecule of water, $M$ is the molar mass of water, and$N_A$is the Avogadro’s number.

$\begin{array}{c}q=\frac{\left(0.0001\right)\left(90\text{\hspace{0.17em}kg}\right)(6.023\times {10}^{23})(18)(1.6\times {10}^{-19}\text{\hspace{0.17em}C})}{0.018\text{\hspace{0.17em}kg/mol}}\\ =8.7\times {10}^5\text{C}\end{array}$

Mary has half the charge carried by John. So, the force of attraction between them is given as-

$\begin{array}{c}F=k\frac{q(q/2)}{d^2}\\ =(9\times {10}^9{\text{\hspace{0.17em}Nm}}^{\text{2}}{\text{/C}}^{\text{2}})\frac{{(8.7\times {10}^5\text{\hspace{0.17em}C})}^2}{2{(30\text{m})}^2}\\ =4\times {10}^{18}\text{N}\end{array}$

Thus, the order of magnitude of the electrostatic force is ${10}^{18}$N.