# Finding F(q) for a Uniform Spherical Charge Model of the Proton

• jeebs
In summary, in order to find the expression for F(q) for a simple model of the proton, we use the fact that F(q) is the Fourier transform of the normalised charge distribution. After making the necessary approximations, we are able to demonstrate that in the limit \frac{qR}{\hbar} << 1, the form factor reduces to 1. However, in order to obtain this result, it is important to use 2nd order approximations for the sine and cosine functions.
jeebs
Here is the problem:

Use the fact that the form factor, F(q), is the Fourier transform of the normalised charge
distribution $$\rho$$(r), which in the spherically symmetric case gives

$$F(q) = \int \frac{4\pi \hbar r}{q}\rho(r) sin(qr/\hbar))dr$$

to find an expression for F(q) for a simple model of the proton considered as a uniform
spherical charge distribution of radius R.

Using your calculated expression for F(q), demonstrate that in the limit
$$\frac{qR}{\hbar} << 1$$
the form factor reduces to 1.

So, what I have tried so far:

I said that for 0 < r < R, the charge density is constant, and could be taken outside the integral along with the other constants, leaving me with

$$F(q) = \frac{4\pi \hbar \rho}{q}\int^R_0 r.sin(qr/\hbar)dr$$

which when I integrate by parts leads to

$$F(q) = \frac{4\pi \hbar^3 \rho}{q^3}sin(qR/\hbar) - \frac{4\pi \hbar^2 \rho R}{q^2}cos(qR/\hbar)$$

I am certain that I have done the integration correctly.

Then I come to the part where I make the approximation that $$\frac{qR}{\hbar} << 1$$ and this happens:
$$sin(qR/\hbar) \approx qR/\hbar$$

and

$$cos(qR/\hbar) \approx 1$$

which gives me

$$F(q) = \frac{4\pi \hbar^2 \rho R}{q^2} - \frac{4\pi \hbar^2 \rho R}{q^2} = 0$$

but I am supposed to be getting F(q) = 1 when I make this approximation.

What am I doing wrong here? My only thought was that the question mentions something about normalization, and I thought that may have something to do with something, but I couln't think what. Any suggestions?

Thanks.

I used a charge density of e/volume (3 e /4 pi R^3) integrated over all space in spherical polars & normalised it then put it in the equation. It basically just got rid of the e

But i still get 0 for the 2nd bit!
edit: Hooray, i get 1 using 2nd order approximations for sin & cos

Last edited:

## 1. How is F(q) calculated for a Uniform Spherical Charge Model of the Proton?

F(q) is calculated using the Coulomb's Law equation, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

## 2. What is the significance of using a Uniform Spherical Charge Model for the Proton?

The Uniform Spherical Charge Model is a simplified representation of the proton that assumes a uniform distribution of charge throughout the proton. This allows for easier calculations and provides a good approximation of the proton's behavior in many situations.

## 3. How does F(q) change as the distance from the proton increases?

As the distance from the proton increases, the force of attraction or repulsion between the proton and another charged particle decreases. This is because the strength of the force is inversely proportional to the square of the distance between the particles.

## 4. Does the charge of the proton affect the value of F(q)?

Yes, the charge of the proton directly affects the value of F(q). As the charge of the proton increases, the force between the proton and another charged particle also increases.

## 5. Can F(q) be used to accurately predict the behavior of a proton in all situations?

No, F(q) is based on the Uniform Spherical Charge Model, which is a simplified representation of the proton. In reality, the proton has a more complex structure and its behavior may not always follow the predictions of F(q).

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