91Ó°ÊÓ

Force exerting by O-H on O, let the left direction be positive and the right direction be negative.

Force between Oxygen and Hydrogen $F_{O-H}$is

$$\begin{gathered} F_{O−H}=\frac{K{e}^2}{R_{O−H}^2} \\ =\frac{K{e}^2}{{\left(R_{O−O}−R_{O−H}\right)}^2} \\ =\frac{\left(9×{10}^9\right){\left(1.6×{10}^{−19}\right)}^2}{{\left[(0.29−0.11)×{10}^{−9}\right]}^2} \\ =7.11×{10}^{−9}\mathrm{N} \end{gathered}$$

Therefore, force between Oxygen and Oxygen $F_{O-O}$is

$$\begin{gathered} F_{O−O}=\frac{K{e}^2}{R_{O−O}^2} \\ =\frac{\left(9×{10}^9\right){\left(1.6×{10}^{−19}\right)}^2}{{\left[0.29×{10}^{−9}\right]}^2} \\ =−2.74×{10}^{−9}\mathrm{N} \end{gathered}$$

Force exerting by H-N on N, let the left direction be positive and the right direction be negative.

Therefore, the force between Hydrogen and Nitrogen is $F_{N-H}$is

$$\begin{gathered} F_{N−H}=\frac{K{e}^2}{R_{N−H}^2} \\ =\frac{K{e}^2}{{\left(R_{N−N}−R_{N−H}\right)}^2} \\ =\frac{\left(9×{10}^9\right){\left(1.6×{10}^{−19}\right)}^2}{{\left[(0.3−0.11)×{10}^{−9}\right]}^2} \\ =6.38×{10}^{−9}\mathrm{N} \end{gathered}$$

The force between Nitrogen and Nitrogen is

$$\begin{gathered} F_{N−N}=\frac{K{e}^2}{R_{N−N}^2} \\ =\frac{\left(9×{10}^9\right){\left(1.6×{10}^{−19}\right)}^2}{{\left[0.3×{10}^{−9}\right]}^2} \\ =−2.56×{10}^{−9}\mathrm{N} \end{gathered}$$

As the third part is the same as the first part with the same force$F_{O-H}$

So, the total net force is

$$\begin{gathered} F_{net}=(2×7.11−2×2.74+6.38−2.56)×{10}^{−9} \\ =1.26×{10}^{−8}N \end{gathered}$$

Hence, Total net force by cytosine exerting by guanine is $1.26×{10}^{-8}N$and is an attractive force.