- #1
Jonsson
- 78
- 0
Hello there,
Imagine that a nucleus consists of three atoms arranged in a equilateral triangle with the length of each side, ##R=2 \rm fm##.
Our protons starts infinitely far away. What is the work required to push these protons together in order to overcome the electric force between them?
I imagine that at any point in time, the force between each pair of protons is given by:
$$
F_E = k\frac{q_0^2}{r^2}
$$
Since each proton sees force from two protons, each at an angle ##\theta = 60^\circ## relative to the other since the triangle has angles $60^\circ$. The magnitude of the sum of these two force vectors must be given by:
$$
\|\mathbf{F}\| = 2\cdot\cos(\theta/2)F_E = 3^{1/2}k\frac{q_0^2}{r^2}
$$
We can then compute the work by letting ##F_{net} = 3 \, \|\mathbf{F}\|##:
$$
W_{\infty\to R} = \int_\infty^R\,F_{net}\,{\rm d}l = 3^{3/2}kq_0\int_\infty^R \frac{1}{r^2}\,{\rm d}l
$$
which is incorrect.
What of the previous is incorrect?
Thank you for your time.
Kind regards,
Marius
Imagine that a nucleus consists of three atoms arranged in a equilateral triangle with the length of each side, ##R=2 \rm fm##.
Our protons starts infinitely far away. What is the work required to push these protons together in order to overcome the electric force between them?
I imagine that at any point in time, the force between each pair of protons is given by:
$$
F_E = k\frac{q_0^2}{r^2}
$$
Since each proton sees force from two protons, each at an angle ##\theta = 60^\circ## relative to the other since the triangle has angles $60^\circ$. The magnitude of the sum of these two force vectors must be given by:
$$
\|\mathbf{F}\| = 2\cdot\cos(\theta/2)F_E = 3^{1/2}k\frac{q_0^2}{r^2}
$$
We can then compute the work by letting ##F_{net} = 3 \, \|\mathbf{F}\|##:
$$
W_{\infty\to R} = \int_\infty^R\,F_{net}\,{\rm d}l = 3^{3/2}kq_0\int_\infty^R \frac{1}{r^2}\,{\rm d}l
$$
which is incorrect.
What of the previous is incorrect?
Thank you for your time.
Kind regards,
Marius
Last edited:
