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| <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> | | <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span> |
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| <p class="author"> | | <p class="author"> |
| Solutions of Savchenko Problems in Physics <br> | | Solutions of Savchenko Problems in Physics <br> |
| <i><b>knowledge must be free</b></i> | | <i><b>knowledge must be free</b></i> |
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| <h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3> | | <h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3> |
| | | |
| <h3> Statement </h3> | | <h3> Statement </h3> |
| <p> | | <p> |
| $6.2.13.$ In a uniformly charged ball of radius $R$, a spherical cavity of radius $r$ was cut out, the center of which is located at a distance $l$ from the center of the ball. Bulk charge density is $\rho$. Find the electric field strength along a straight line passing through the center of the cavity and the center of the ball. Prove that the electric field in the cavity is homogeneous. | | $6.2.13.$ In a uniformly charged ball of radius $R$, a spherical cavity of radius $r$ was cut out, the center of which is located at a distance $l$ from the center of the ball. Bulk charge density is $\rho$. Find the electric field strength along a straight line passing through the center of the cavity and the center of the ball. Prove that the electric field in the cavity is homogeneous. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="statement.png" | | <img src="statement.png" |
| loading="lazy" width="350" /> | | loading="lazy" width="350" /> |
| <figcaption> | | <figcaption> |
| For problem $6.2.13$ | | For problem $6.2.13$ |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <h3>Solution</h3> | | <h3>Solution</h3> |
| <p> | | <p> |
| Here, the idea for solving the problem is to use superposition's principle: considering a sphere of radius $R$ (bigger one) with bulk charge density $\rho$ and smaller one with bulk charge density $-\rho$, such that when superposition occurs, the cavity have no charge. Moreover, let's consider the following figure for organizing the solving process. | | Here, the idea for solving the problem is to use superposition's principle: considering a sphere of radius $R$ (bigger one) with bulk charge density $\rho$ and smaller one with bulk charge density $-\rho$, such that when superposition occurs, the cavity have no charge. Moreover, let's consider the following figure for organizing the solving process. |
| </p> | | </p> |
| <center> | | <center> |
| <figure> | | <figure> |
| <img src="analysis.png" | | <img src="analysis.png" |
| loading="lazy" width="200" /> | | loading="lazy" width="200" /> |
| <figcaption> | | <figcaption> |
| Regions for analysis | | Regions for analysis |
| </figcaption> | | </figcaption> |
| </figure> | | </figure> |
| </center> | | </center> |
| <p> | | <p> |
| Above picture shows four regions to study. Let's begin with first one... Applying Gauss's Law for bigger sphere (without cavity) at distance $x$ from its center, | | Above picture shows four regions to study. Let's begin with first one... Applying Gauss's Law for bigger sphere (without cavity) at distance $x$ from its center, |
| $$E_b(x)\cdot 4\pi x^2 = \frac{q_{enc}}{\varepsilon_0} = \frac{\rho\cdot 4\pi x^3}{3\varepsilon_0}$$ | | $$E_b(x)\cdot 4\pi x^2 = \frac{q_{enc}}{\varepsilon_0} = \frac{\rho\cdot 4\pi x^3}{3\varepsilon_0}$$ |
| $$E_b(x) = \frac{\rho x}{3\varepsilon_0} \;(1)$$ | | $$E_b(x) = \frac{\rho x}{3\varepsilon_0} \;(1)$$ |
| while for smaller one, | | while for smaller one, |
| $$E_s(r')\cdot 4\pi r'^2 = \frac{q_{enc}'}{\varepsilon_0} = \frac{\rho\cdot 4\pi r^3}{3\varepsilon_0}$$ | | $$E_s(r')\cdot 4\pi r'^2 = \frac{q_{enc}'}{\varepsilon_0} = \frac{\rho\cdot 4\pi r^3}{3\varepsilon_0}$$ |
| $$E_s(r') = \frac{\rho r^3}{3 r'^2 \varepsilon_0} \;(2)$$ | | $$E_s(r') = \frac{\rho r^3}{3 r'^2 \varepsilon_0} \;(2)$$ |
| In this region $\vec{E_b}\uparrow\uparrow \vec{E_s}$, overlapping | | In this region $\vec{E_b}\uparrow\uparrow \vec{E_s}$, overlapping |
| $$E(x,r') = E_s(r') + E_b(x) \;(3)$$ | | $$E(x,r') = E_s(r') + E_b(x) \;(3)$$ |
| Putting (1) and (2) into (3), | | Putting (1) and (2) into (3), |
| $$E(x,r') = \frac{\rho}{3\varepsilon_0}\left(x+\frac{r^3}{r'^2}\right)$$ | | $$E(x,r') = \frac{\rho}{3\varepsilon_0}\left(x+\frac{r^3}{r'^2}\right)$$ |
| but $r' = l-x$, so | | but $r' = l-x$, so |
| </p> | | </p> |
| <h4>Answer 1</h4> | | <h4>Answer 1</h4> |
| <p> | | <p> |
| $$E_I(x) = \frac{\rho}{3\varepsilon_0}\left[x+\frac{r^3}{(l-x)^2}\right]$$ | | $$E_I(x) = \frac{\rho}{3\varepsilon_0}\left[x+\frac{r^3}{(l-x)^2}\right]$$ |
| for $0\leq x\leq l-r$. | | for $0\leq x\leq l-r$. |
| </p> | | </p> |
| | | |
| <p> | | <p> |
| For region II, let's separate region in two subregions II$_1$ and II$_2$. For first subregion, field generated by bigger sphere (without cavity) is given by (1). While for smaller one, | | For region II, let's separate region in two subregions II$_1$ and II$_2$. For first subregion, field generated by bigger sphere (without cavity) is given by (1). While for smaller one, |
| $$E_s(r')\cdot 4\pi r'^2 = \frac{q_{enc}'}{\varepsilon_0} = \frac{\rho\cdot 4\pi r'^3}{3\varepsilon_0}$$ | | $$E_s(r')\cdot 4\pi r'^2 = \frac{q_{enc}'}{\varepsilon_0} = \frac{\rho\cdot 4\pi r'^3}{3\varepsilon_0}$$ |
| $$E_s(r') = \frac{\rho r'}{3\varepsilon_0} \;(4)$$ | | $$E_s(r') = \frac{\rho r'}{3\varepsilon_0} \;(4)$$ |
| In this subregion, $\vec{E_s}\uparrow\uparrow \vec{E_b}$, so overlapping | | In this subregion, $\vec{E_s}\uparrow\uparrow \vec{E_b}$, so overlapping |
| $$E_{II_1}(x,r') = E_s(r') + E_b(x) \;(5)$$ | | $$E_{II_1}(x,r') = E_s(r') + E_b(x) \;(5)$$ |
| Substituting (1) and (4) into (5), | | Substituting (1) and (4) into (5), |
| $$E_{II_1}(x,r') = \frac{\rho}{3\varepsilon_0}(x+r')$$ | | $$E_{II_1}(x,r') = \frac{\rho}{3\varepsilon_0}(x+r')$$ |
| but $l=x+r'$, hence | | but $l=x+r'$, hence |
| $$E_{II_1} = \frac{\rho l}{3\varepsilon_0}$$ | | $$E_{II_1} = \frac{\rho l}{3\varepsilon_0}$$ |
| For second subregion, field generated by bigger sphere (without cavity) still given by (1) and the generated one by smaller sphere is given by (4), but $\vec{E_s}\uparrow\downarrow\vec{E_b}$, so overlapping | | For second subregion, field generated by bigger sphere (without cavity) still given by (1) and the generated one by smaller sphere is given by (4), but $\vec{E_s}\uparrow\downarrow\vec{E_b}$, so overlapping |
| $$E_{II_2}(x,r') = E_b(x) - E_s(r') \;(6)$$ | | $$E_{II_2}(x,r') = E_b(x) - E_s(r') \;(6)$$ |
| Putting (1) and (4) into (6) | | Putting (1) and (4) into (6) |
| $$E_{II_2}(x,r') = \frac{\rho}{3\varepsilon_0}(x-r')$$ | | $$E_{II_2}(x,r') = \frac{\rho}{3\varepsilon_0}(x-r')$$ |
| but in this case $l = x-r'$, | | but in this case $l = x-r'$, |
| $$E_{II_2} = \frac{\rho l}{3\varepsilon_0}$$ | | $$E_{II_2} = \frac{\rho l}{3\varepsilon_0}$$ |
| Finally, $E_{II_1} = E_{II_2} = E_{II}$ | | Finally, $E_{II_1} = E_{II_2} = E_{II}$ |
| </p> | | </p> |
| <h4>Answer 2</h4> | | <h4>Answer 2</h4> |
| <p> | | <p> |
| $$E_{II} = \frac{\rho l}{3\varepsilon_0}$$ | | $$E_{II} = \frac{\rho l}{3\varepsilon_0}$$ |
| for $l-r\leq x \leq l+r$ | | for $l-r\leq x \leq l+r$ |
| </p> | | </p> |
| | | |
| <p> | | <p> |
| For region III, smaller-sphere's field is given by (2), while bigger-sphere's field is given by (1), as $\vec{E_s}\uparrow\downarrow\vec{E_b}$, overlapping | | For region III, smaller-sphere's field is given by (2), while bigger-sphere's field is given by (1), as $\vec{E_s}\uparrow\downarrow\vec{E_b}$, overlapping |
| $$E_{III}(r',x) = E_b(x) - E_s(r') \;(7)$$ | | $$E_{III}(r',x) = E_b(x) - E_s(r') \;(7)$$ |
| Putting (1) and (2) into (7) | | Putting (1) and (2) into (7) |
| $$E_{III}(x,r') = \frac{\rho}{3\varepsilon_0}\left(x - \frac{r^3}{r'^2}\right)$$ | | $$E_{III}(x,r') = \frac{\rho}{3\varepsilon_0}\left(x - \frac{r^3}{r'^2}\right)$$ |
| as $r' = x-l$, | | as $r' = x-l$, |
| </p> | | </p> |
| <h4>Answer 3</h4> | | <h4>Answer 3</h4> |
| <p> | | <p> |
| $$E_{III}(x) = \frac{\rho}{3\varepsilon_0}\left[x-\frac{r^3}{(x-l)^2}\right]$$ | | $$E_{III}(x) = \frac{\rho}{3\varepsilon_0}\left[x-\frac{r^3}{(x-l)^2}\right]$$ |
| for $l+r \leq x \leq R$. | | for $l+r \leq x \leq R$. |
| </p> | | </p> |
| | | |
| <p> | | <p> |
| For region IV (points outside bigger sphere), applying Gauss's Law for bigger sphere, | | For region IV (points outside bigger sphere), applying Gauss's Law for bigger sphere, |
| $$E_b(x) \cdot 4\pi x^2 = \frac{q_{enc}}{\varepsilon_0} = \frac{\rho\cdot 4\pi R^3}{3\varepsilon_0}$$ | | $$E_b(x) \cdot 4\pi x^2 = \frac{q_{enc}}{\varepsilon_0} = \frac{\rho\cdot 4\pi R^3}{3\varepsilon_0}$$ |
| $$E_b(x) = \frac{\rho R^3}{3\varepsilon_0 x^2} \;(8)$$ | | $$E_b(x) = \frac{\rho R^3}{3\varepsilon_0 x^2} \;(8)$$ |
| while for smaller one, field is given by (4). Since $\vec{E_s}\uparrow\downarrow\vec{E_b}$, overlapping | | while for smaller one, field is given by (4). Since $\vec{E_s}\uparrow\downarrow\vec{E_b}$, overlapping |
| $$E_{IV}(x,r') = E_b(x) - E_s(r') \;(9)$$ | | $$E_{IV}(x,r') = E_b(x) - E_s(r') \;(9)$$ |
| Putting (4) and (8) into (9) | | Putting (4) and (8) into (9) |
| $$E_{IV}(x,r') = \frac{\rho}{3\varepsilon_0}\left(\frac{R^3}{x^2}-\frac{r^3}{r'^2}\right)$$ | | $$E_{IV}(x,r') = \frac{\rho}{3\varepsilon_0}\left(\frac{R^3}{x^2}-\frac{r^3}{r'^2}\right)$$ |
| but $l = x-r'$, | | but $l = x-r'$, |
| </p> | | </p> |
| <h4>Answer 4</h4> | | <h4>Answer 4</h4> |
| <p> | | <p> |
| $$E_{IV}(x) = \frac{\rho}{3\varepsilon_0}\left[\frac{R^3}{x^2}-\frac{r^3}{(x-l)^2}\right]$$ | | $$E_{IV}(x) = \frac{\rho}{3\varepsilon_0}\left[\frac{R^3}{x^2}-\frac{r^3}{(x-l)^2}\right]$$ |
| for $x > R$. | | for $x > R$. |
| </p> | | </p> |
| <p> | | <p> |
| For proofing the homogenity of field inside cavity, let's consider any point P inside cavity. Absolute value of bigger-sphere's field is given by (1) and directed over $\hat{x}$ direction, while absolute value of smaller-sphere's field is given by (4) and directed over $-\hat{r'}$ direction. Overlapping | | For proofing the homogenity of field inside cavity, let's consider any point P inside cavity. Absolute value of bigger-sphere's field is given by (1) and directed over $\hat{x}$ direction, while absolute value of smaller-sphere's field is given by (4) and directed over $-\hat{r'}$ direction. Overlapping |