Answer
The magnitude of the net electric force exerted by one base on the other is $3.99 \times 10^{-10} N$
Work Step by Step
The values we have are:
$k = 8.99 \times 10^9\ Nm^2/C^2$
$N = (-0.3)(1.602 \times 10^{-19}\ C) = -4.806 \times 10^{-20}\ C$
$H = (0.3)(1.602 \times 10^{-19}\ C) = 4.806 \times 10^{-20}\ C$
$O = (-0.4)(1.602 \times 10^{-19}\ C) = -6.408 \times 10^{-20}\ C$
$C = (0.4)(1.602 \times 10^{-19}\ C) = 6.408 \times 10^{-20}\ C$
Now we calculate the net force for each charge:
$F_{NO} = \dfrac{k*N*O}{L^2} = \dfrac{(8.99 \times 10^9\ Nm^2/C^2)(-4.806 \times 10^{-20}\ C)(-6.408 \times 10^{-20}\ C)}{(3\times 10^{-10}\ m )^2}
= -3.08 \times 10^{-10}\ N$
$F_{NC} = \dfrac{k*N*C}{L^2} = \dfrac{(8.99 \times 10^9\ Nm^2/C^2)(-4.806 \times 10^{-20}\ C)(6.408 \times 10^{-20}\ C)}{(4.2\times 10^{-10}\ m )^2} = 1.57 \times 10^{-10}\ N$
$F_{HO} = \dfrac{k*H*O}{L^2} = \dfrac{(8.99 \times 10^9\ Nm^2/C^2)(4.806 \times 10^{-20}\ C)(-6.408 \times 10^{-20}\ C)}{(1.8\times 10^{-10}\ m )^2} = 8.55 \times 10^{-10}\ N$
$F_{HC} = \dfrac{k*H*C}{L^2} = \dfrac{(8.99 \times 10^9\ Nm^2/C^2)(4.806 \times 10^{-20}\ C)(6.408 \times 10^{-20}\ C)}{(3\times 10^{-10}\ m )^2} = -3.08 \times 10^{-10}\ N$
We add up forces:
$\sum F = -3.08 \times 10^{-10}\ N + 1.57 \times 10^{-10}\ N + 8.55 \times 10^{-10}\ N - 3.08 \times 10^{-10}\ N = 3.96 \times 10^{-10}\ N $
The magnitude of the net electric force exerted by one base on the other is $3.99 \times 10^{-10} N$.
