What Is the Net Electric Force in a DNA Hydrogen Bond?

AI Thread Summary
The discussion focuses on calculating the net electric force in a DNA hydrogen bond, specifically between adenine and thymine, modeled as four point charges. Initial calculations using incorrect charge values led to confusion, prompting a reevaluation of the charge magnitudes based on the charge of an electron. After correcting the values, the participant recalculated the forces between the pairs and found the correct net force. The final result for the net electric force was determined to be approximately 3.99x10^-10 N. This highlights the importance of accurate charge values in electrostatic calculations.
StormPix
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Homework Statement



In a DNA molecule, the base pair adenine and thymine is held together by two hydrogen bonds (see figure below).

16-figure-05.gif


Let's model one of these hydrogen bonds as four point charges arranged along a straight line. Using the information in the figure below, calculate the magnitude of the net electric force along this hydrogen bond.

16-p-019.gif


Homework Equations



F=(k*|q1|*|q2|)/r^2

The Attempt at a Solution



Since we are looking at 4 point charges, we need 4 equations. They are:

k = 8.99x10^9 (N*m^2)/C^2
N = -3x10^-10 C
O = -4x10^-10 C
H = 3x10^-10 C
C = 4x10^-10 C

F_no = (k*|N|*|O|) / ((3x10^-10 m)^2) = -1.20x10^10 N -> negative as repelling force
F_nc = (k*|N|*|C|) / ((4.2x10^-10 m)^2) = 6.12x10^9 N
F_ho = (k*|H|*|O|) / ((1.8x10^-10 m)^2) = 3.33x10^10 N
F_hc = (k*|H|*|C|) / ((3x10^-10 m)^2) = -1.2x10^10 N -> negative as repelling force

Since we are asked for find the magnitude of the net electric force, we need to sum all 4 forces together.

ΣF = F_no + F_nc + F_ho + F_hc = 1.54x10^10 N

... Opps. So I must have done something wrong.

The hint given was "You need 4 eqns: pair N&O, N&C, H&O, H&C and use the charges & distances appropriate for each pair to get a net magnitude between the bases." ... which to my knowledge I have here.

So not sure what I did wrong. Any help would be much appreciated.
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StormPix said:
N = -3x10^-10 C
O = -4x10^-10 C
H = 3x10^-10 C
C = 4x10^-10 C

Check these numbers. What does "e" stand for in the picture?
 
I'm not following your arithmetic.
It would be clearer to leave out k and the unit charge of an electron as common factors to bring in later. E.g. for N::O write -0.3*0.4/(0.12+0.18)2.
 
TSny said:
Check these numbers. What does "e" stand for in the picture?

Ah! Small but critical oversight on my part. "e" meant electron ... so that each base pair is a factor .3 or .4 charge of an electron.

This would make:

N = (-.3)(1.602x10^-19 C) = -4.806x10^-20 C
H = (.3)(1.602x10^-19 C) = 4.806x10^-20 C
O = (-.4)(1.602x10^-19 C) = -6.408x10^-20 C
C = (.4)(1.602x10^-19 C) = 6.408x10^-20 C

Using these new values for N, H, O, C I get

F_no = -3.08x10^-10 N -> repelling force
F_nc = 1.57x10^-10 N
F_ho = 8.55x10^-10 N
F_hc = -3.08x10^-10 N -> repelling force

Such that,

ΣF = 2*(-3.08x10^-10 N) + 1.57x10^-10 N + 8.55x10^-10 N = 3.99x10^-10 N which ended up be the correct answer.

Thanks a lot.
 
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