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Found in: Page 1251

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

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# In 1911, Ernest Rutherford modeled an atom as being a point of positive charge surrounded by a negative charge -ze uniformly distributed in a sphere of radius centered at the point. At distance within the sphere, the electric potential is ${\mathbf{V}}{\mathbf{=}}\frac{\mathbf{Ze}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}}\left(\frac{1}{r}-\frac{3}{2R}+\frac{{r}^{2}}{2{R}^{3}}\right)$ .From this formula, determine the magnitude of the electric field for ${\mathbf{0}}{\mathbf{\le }}{\mathbf{r}}{\mathbf{\le }}{\mathbf{R}}$. What are the (b) electric field and (c) potential for ${\mathbf{r}}{\mathbf{\ge }}{\mathbf{R}}$ ?

1. The magnitude of the electric field for $0\le \mathrm{r}\le \mathrm{R}$ is $\frac{\mathrm{Ze}}{4{\mathrm{\pi \epsilon }}_{0}}\left(\frac{1}{{\mathrm{r}}^{2}}-\frac{\mathrm{r}}{{\mathrm{R}}^{3}}\right)$.
2. The electric field for $\mathrm{r}\ge \mathrm{R}$ is 0.
3. The potential for $\mathrm{r}\ge \mathrm{R}$ is 0.
See the step by step solution

## Step 1: The given data

1. A positive charge Ze is surrounded by a -Ze negative charge uniformly distributed in a sphere of radius R.
2. The electric potential at a distance r within the sphere, $\mathrm{V}=\frac{\mathrm{Ze}}{4{\mathrm{\pi \epsilon }}_{0}}\left(\frac{1}{\mathrm{r}}-\frac{3}{2\mathrm{R}}+\frac{{\mathrm{r}}^{3}}{2{\mathrm{R}}^{3}}\right)$

## Step 2: Understanding the concept of electric field and potential:

Electric field is the negative space derivation of electric potential.

Using the given formula in the electric field equation, get the magnitude of the electric field by differentiating the given equation. Now, substituting the value of for , to get the value of both the electric field and potential at that point. Again, as there is no charge outside a uniformly charged sphere, its potential is zero outside the sphere surface.

Formula:

The relation of electric field due to charge and changing potential is,

$\mathrm{E}=-\frac{\partial \mathrm{V}}{\partial \mathrm{r}}$ ….. (1)

## Step 3: (a) Calculation of the magnitude of the electric field:

Using the given potential equation in equation (1), the magnitude of the electric field for the condition $0\le \mathrm{r}\le \mathrm{R}$ can be given as follows:

$\mathrm{E}=-\frac{\partial }{\partial \mathrm{r}}\left[\frac{\mathrm{Ze}}{4{\mathrm{\pi \epsilon }}_{0}}\left(\frac{1}{\mathrm{r}}-\frac{3}{2\mathrm{R}}+\frac{{\mathrm{r}}^{3}}{2{\mathrm{R}}^{3}}\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{E}=\frac{\mathrm{Ze}}{4{\mathrm{\pi \epsilon }}_{0}}\left(\frac{1}{{\mathrm{r}}^{2}}-\frac{\mathrm{r}}{{\mathrm{R}}^{3}}\right)$ ….. (2)

Hence, the value of the electric field is $\frac{\mathrm{Ze}}{4{\mathrm{\pi \epsilon }}_{0}}\left(\frac{1}{{\mathrm{r}}^{2}}-\frac{\mathrm{r}}{{\mathrm{R}}^{3}}\right)$ .

## Step 4: (b) Calculation of the electric field at r≥R :

Now, the electric field at r = R is given using equation (2) as follows:

${\mathrm{E}}_{\mathrm{r}=\mathrm{R}}=\frac{\mathrm{Ze}}{4{\mathrm{\pi \epsilon }}_{0}}\left(\frac{1}{{\mathrm{R}}^{2}}-\frac{\mathrm{R}}{{\mathrm{R}}^{3}}\right)\phantom{\rule{0ex}{0ex}}=0$

Thus, the field vanishes at r = R .

Since, the value of potential outside the sphere is , V = 0 this conclude that the electric field is also zero considering equation (1).

Hence, the value of the electric field for $\mathrm{r}\ge \mathrm{R}$ is 0 .

## Step 5: (c) Calculation of the potential at :

Now, the potential at outside the sphere is V = 0 .

So, the value of potential at is given as follows:

${\mathrm{V}}_{\mathrm{r}=\mathrm{R}}=\frac{\mathrm{Ze}}{4{\mathrm{\pi \epsilon }}_{0}}\left(\frac{1}{\mathrm{R}}-\frac{3}{2\mathrm{R}}+\frac{{\mathrm{R}}^{3}}{2{\mathrm{R}}^{3}}\right)$

Hence, the value of the potential for $\mathrm{r}\ge \mathrm{R}$ is 0.

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