Calculate binding energy & compare to Electrostatic Repulsion

[ZedCar]

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]

Which would be;

(1.6 x 10^-19)^2 / [4∏ x 8.85 x 10^-12 x (1 x 10^-15)^2]

= 230 J

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.

So I wasn't sure which is correct. Thanks for any advice!

 

[collinsmark]

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?

 

[ZedCar]

So is the problem statement asking for electrostatic energy or force?

Thanks very much collinsmark.

The units of the answer given in my notes has units of Joules. So the question would appear to be asking for electrostatic energy.

So the equation they have used appears to be correct.