# Electric force in DNA molecule

Given that distance between O-H and N-H bonds are 0.11 nm and

How do I compute the net force exerted from Thymine and Adenine? A hint given is:

Hint: To keep calculations fairly simple, yet reasonable, consider only the forces due to the O–H–N and the N–H–N combinations, assuming that these two combinations are parallel to each other.

Do I compute the force on O due to H and N like:

$$F_{OH} = k \frac{e^2}{(0.17 \times 10^{-9})^2}$$

$$F_{ON} = k \frac{e^2}{(0.28 \times 10^{-9})^2}$$

$$F_{O} = k \frac{e^2}{(0.17 \times 10^{-9})^2} - k \frac{e^2}{(0.28 \times 10^{-9})^2}$$

Then about the HN part? Do I consider that 1 molecule or something?

-
that looks pretty good. assuming you only have to consider the forces due to the two hydrogen bonds (O~H and N~H), then you've done the O~H part. – Ethan Jan 27 '13 at 8:34
@user1544418, I did do the ON part too? Or do you mean I can now proceed to compute the force on the N (on Adenine) in a similar way and sum up those forces? – Jiew Meng Jan 27 '13 at 9:16
yes exactly. be sure you realize that the forces are symmetric - the force on adenine is the same as that on thymine (this is newtons third law). – Ethan Jan 27 '13 at 9:29