Question

Estimate the net force between the CO group and the HN group shown in Fig. 16–63. The C and O have charges\({\bf{ \pm 0}}{\bf{.40e}}\)and the H and N have charges\({\bf{ \pm 0}}{\bf{.20e}}\), where\({\bf{e = 1}}{\bf{.6 \times 1}}{{\bf{0}}^{{\bf{ - 19}}}}\;{\bf{C}}\). (Hint: Do not include the “internal” forces between C and O, or between H and N.)

FIGURE 16–63 Problem 50

Short answer

The net force between the CO group and the NH group is \(2.4 \times {10^{ - 10}}\;{\rm{N}}\).

Step-by-step solution

Understanding the electric force

A point charge exerts an electric force on another point charge in the vicinity. It could be attractive as well as repulsive depending upon the nature of the two charges.

The expression for the electric force is given as:

\(F = k\frac{{{Q_1}{Q_2}}}{{{r^2}}}\) … (i)

Here, k is the Coulomb’s constant,\({Q_1},\;{Q_2}\)are the charges and r is the separation between the charges.

Given Data

The charge of C and O is, \({q_1} = \pm 0.4e\).

The charge of H and N is, \({q_2} = \pm 0.2e\).

The electronic charge is, \(e = 1.6 \times {10^{ - 19}}\;{\rm{C}}\).

The distance between C and O is, \({r_1} = 0.12\;{\rm{nm}}\).

The distance between N and H is, \({r_2} = 0.10\;{\rm{nm}}\).

The distance between N and O is, \({r_3} = 0.28\;{\rm{nm}}\).

Determination of net force between the CO and NH group

The expression to determine the net force is given as:

\(\begin{aligned}{c}{F_{\rm{n}}} &= {F_{{\rm{CH}}}} + {F_{{\rm{CN}}}} + {F_{{\rm{OH}}}} + {F_{{\rm{ON}}}}\\{F_{\rm{n}}} &= - \frac{{k{q_1}{q_2}}}{{{{\left( {{r_1} + \left( {{r_3} - {r_2}} \right)} \right)}^2}}} + \frac{{k{q_1}{q_2}}}{{{{\left( {{r_1} + {r_3}} \right)}^2}}} + \frac{{k{q_1}{q_2}}}{{{{\left( {{r_3} - {r_2}} \right)}^2}}} - \frac{{k{q_1}{q_2}}}{{{r_3}^2}}\\{F_{\rm{n}}} &= \left( {k{q_1}{q_2}} \right)\left( { - \frac{1}{{{{\left( {{r_1} + \left( {{r_3} - {r_2}} \right)} \right)}^2}}} + \frac{1}{{{{\left( {{r_1} + {r_3}} \right)}^2}}} + \frac{1}{{{{\left( {{r_3} - {r_2}} \right)}^2}}} - \frac{1}{{{r_3}^2}}} \right)\end{aligned}\)

Here, kis Coulomb’s constant.

Substitute the values in the above expression.

\(\begin{aligned}{l}{F_{\rm{n}}} &= \left( \begin{aligned}{l}\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right) \times \\\left( {0.4 \times 1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right) \times \\\left( {0.2 \times 1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)\end{aligned} \right)\left( \begin{aligned}{l} - \frac{1}{{{{\left( {\left( {0.12 + \left( {0.28 - 0.10} \right)} \right) \times {{10}^{ - 9}}\;{\rm{m}}} \right)}^2}}} + \\\frac{1}{{{{\left( {\left( {0.12 + 0.28} \right) \times {{10}^{ - 9}}\;{\rm{m}}} \right)}^2}}}\\ + \frac{1}{{{{\left( {\left( {0.28 - 0.10} \right) \times {{10}^{ - 9}}\;{\rm{m}}} \right)}^2}}} - \frac{1}{{{{\left( {0.28 \times {{10}^{ - 9}}\;{\rm{m}}} \right)}^2}}}\end{aligned} \right)\\{F_{\rm{n}}} &= 2.4 \times {10^{ - 10}}\;{\rm{N}}\end{aligned}\)

Thus, the net force between the CO group and the NH group is \(2.4 \times {10^{ - 10}}\;{\rm{N}}\).