Question

How much work would it take to push two protons very slowly from a separation of 2.00×10−10 m (a typical atomic distance) to 3.00×10−15 m (a typical nuclear distance)?

Answer 1

We can visualize the problem in another way, which is equivalent but easier to solve: let's imagine we hold one proton in the same place, and we move the other proton from a distance of 2.00×10−10 m to a distance of 3.00×10−15 m from the first proton. How much work is done?

The work done is equal to the electric potential energy gained by the proton:

$W=q \Delta V$

where $q=1.6 \cdot 10^{-19}C$ is the charge of the proton and $\Delta V$ is the potential difference between the final position and the initial position of the proton. To calculate this $\Delta V$, we must calculate the electric potential generated by the proton at rest at the two points, using the formula:

$V=k\frac{Q}{r}$

where $k=9.0 \cdot 10^9 N m^2 C^{-2}$ is the Coulomb constant and Q is the proton charge. Substituting the initial and final distance of the second proton, we find

$V_i = (9.0 \cdot 10^9 )\frac{1.6 \cdot 10^{-19}}{2.0 \cdot 10^{-10}}=7.2 V$

$V_f = (9.0 \cdot 10^9 )\frac{1.6 \cdot 10^{-19}}{3.0 \cdot 10^{-15}}=4.8 \cdot 10^5 V$

Therefore, the work done is

$W=q \Delta V=(1.6 \cdot 10^{-19}C)(4.8 \cdot 10^5 V-72 V)=7.7 \cdot 10^{-14} J$