- #1

jamba88

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## Homework Statement

Consider a segment of a molecule of DNA that is a coil with an overall length of 2.1*10^-6 m. If the ends of the molecule become singly ionized (one end loosing a single electron and the other end gaining a single electron) the helical molecule acts like a spring and compresses by 1.08% of its original length. Determine the effective spring constant of the molecule.

## Homework Equations

F(elec)= F(spring)

ke* [(Q1*Q2)/r]= k*(change in r)

## The Attempt at a Solution

The DNA will have a charge of +e and one end and -e at the other end which we will plug in for our Q1 and Q2 values.

I'm particularly concerned if I calculated the radius and change in radius correctly:

Calculate change in r:

.0108*(2.1*10^-6 m)= 2.3*10^-8 m

Calculate r:

2.1*10^.6 m- 2.3*10^-8 m= 2.1 *10^-6 m

It seems fishy that my r is the same as my r initial.

Plug into equation: ke* [(Q1*Q2)/r]= k*(change in r)

8.99*10^9 Nm^2/C^2 *[(-1.60*10^-19C)*(+1.60*10^-19C)/ (2.1*10^-6 m)]= k*(2.1*10^-6 m)

k= -4.76*10^-15