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Q79P

Expert-verifiedFound in: Page 1251

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**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****${\mathbf{r}}{\mathbf{\ge}}{\mathbf{R}}$****?**

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

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

**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)

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

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 .

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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