ZedCar said:
Homework Statement
4He2 has size of ~ 2 x 10^-15 m and a mass of 4.0015u. Calculate the binding energy of this nucleus, and compare it to the electrostatic repulsion of its constituents.
Homework Equations
The Attempt at a Solution
For the first part I have calculated B.E. = 4.5 x 10^-12 J This agrees with my notes.
For the second part do I use the equation E = (q1 x q2) / [4∏(ε0)r^2]
Something is not right with that. If
E is supposed to represent energy, you have a extra
r in the equation. If
E is supposed to represent electric field, you have an extra
q in the equation. If the right side of the equation is correct, the left hand side is a measure of force.
Which would be;
(1.6 x 10^-19)^2 / [4∏ x 8.85 x 10^-12 x (1 x 10^-15)^2]
= 230 J
Okay, if you are trying to calculate energy, you haven't used the correct formula (see above).
The reason I ask is because in the notes I have the r = 1 x 10^-15 value is not squared in the same calculation therefore giving an answer of 2.3 x 10^-13 J.
There are different formulas for force and energy. (see below)
So I wasn't sure which is correct. Thanks for any advice!
The
force magnitude between two charges is
F = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^2}
The units are of course units of force such as Newtons (not Joules).
Invoking the definition of work, dW = \vec F \cdot \vec{ds} (in part), we can calculate the electric potential energy. The electric potential energy is the energy that it takes for one charge brought from infinity to some radius,
r away from some other charge. (I'll call this energy
W):
W = -\int_{-\infty}^r \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r^{'2}} dr^' = \frac{1}{4 \pi \varepsilon_0} \frac{q_1 q_2}{r}
Note that the
r in the denominator is not squared after the integration.
W here has units of energy, such as joules.
So is the problem statement asking for electrostatic energy or force?
