Added Luis's English solution of 6.2.13

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+ <meta name="description" content="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.">
+ <meta name="author" content="Aliaksandr Melnichenka">
+ <meta name="date" content="2023-10" scheme="YYYY-MM">
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+ <title>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.</title>
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+ <span><img src="../../img/book.png"></span><span>Savchenko Solutions</span>
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+ Solutions&nbsp;of&nbsp;Savchenko Problems&nbsp;in&nbsp;Physics <br>
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+ <h3 id="back-link"><a href="../../#6.2">$\leftarrow$Back</a></h3>
+
+ <h3> Statement </h3>
+ <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.
+</p>
+<center>
+ <figure>
+ <img src="statement.png"
+ loading="lazy" width="350" />
+ <figcaption>
+ For problem $6.2.13$
+ </figcaption>
+ </figure>
+</center>
+ <h3>Solution</h3>
+ <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.
+ </p>
+ <center>
+ <figure>
+ <img src="analysis.png"
+ loading="lazy" width="200" />
+ <figcaption>
+ Regions for analysis
+ </figcaption>
+ </figure>
+ </center>
+ <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,
+ $$E_b(x)\cdot 4\pi x^2 = \frac{q_{enc}}{\varepsilon_0} = \frac{\rho\cdot 4\pi x^3}{3\varepsilon_0}$$
+ $$E_b(x) = \frac{\rho x}{3\varepsilon_0} \;(1)$$
+ while for smaller one,
+ $$E_s(r')\cdot 4\pi r'^2 = \frac{q_{enc}'}{\varepsilon_0} = \frac{\rho\cdot 4\pi r^3}{3\varepsilon_0}$$
+ $$E_s(r') = \frac{\rho r^3}{3 r'^2 \varepsilon_0} \;(2)$$
+ In this region $\vec{E_b}\uparrow\uparrow \vec{E_s}$, overlapping
+ $$E(x,r') = E_s(r') + E_b(x) \;(3)$$
+ Putting (1) and (2) into (3),
+ $$E(x,r') = \frac{\rho}{3\varepsilon_0}\left(x+\frac{r^3}{r'^2}\right)$$
+ but $r' = l-x$, so
+ </p>
+ <h4>Answer 1</h4>
+ <p>
+ $$E_I(x) = \frac{\rho}{3\varepsilon_0}\left[x+\frac{r^3}{(l-x)^2}\right]$$
+ for $0\leq x\leq l-r$.
+ </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,
+ $$E_s(r')\cdot 4\pi r'^2 = \frac{q_{enc}'}{\varepsilon_0} = \frac{\rho\cdot 4\pi r'^3}{3\varepsilon_0}$$
+ $$E_s(r') = \frac{\rho r'}{3\varepsilon_0} \;(4)$$
+ In this subregion, $\vec{E_s}\uparrow\uparrow \vec{E_b}$, so overlapping
+ $$E_{II_1}(x,r') = E_s(r') + E_b(x) \;(5)$$
+ Substituting (1) and (4) into (5),
+ $$E_{II_1}(x,r') = \frac{\rho}{3\varepsilon_0}(x+r')$$
+ but $l=x+r'$, hence
+ $$E_{II_1} = \frac{\rho l}{3\varepsilon_0}$$
+ 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
+ $$E_{II_2}(x,r') = E_b(x) - E_s(r') \;(6)$$
+ Putting (1) and (4) into (6)
+ $$E_{II_2}(x,r') = \frac{\rho}{3\varepsilon_0}(x-r')$$
+ but in this case $l = x-r'$,
+ $$E_{II_2} = \frac{\rho l}{3\varepsilon_0}$$
+ Finally, $E_{II_1} = E_{II_2} = E_{II}$
+ </p>
+ <h4>Answer 2</h4>
+ <p>
+ $$E_{II} = \frac{\rho l}{3\varepsilon_0}$$
+ for $l-r\leq x \leq l+r$
+ </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
+ $$E_{III}(r',x) = E_b(x) - E_s(r') \;(7)$$
+ Putting (1) and (2) into (7)
+ $$E_{III}(x,r') = \frac{\rho}{3\varepsilon_0}\left(x - \frac{r^3}{r'^2}\right)$$
+ as $r' = x-l$,
+ </p>
+ <h4>Answer 3</h4>
+ <p>
+ $$E_{III}(x) = \frac{\rho}{3\varepsilon_0}\left[x-\frac{r^3}{(x-l)^2}\right]$$
+ for $l+r \leq x \leq R$.
+ </p>
+
+ <p>
+ For region IV (points outside bigger sphere), applying Gauss's Law for bigger sphere,
+ $$E_b(x) \cdot 4\pi x^2 = \frac{q_{enc}}{\varepsilon_0} = \frac{\rho\cdot 4\pi R^3}{3\varepsilon_0}$$
+ $$E_b(x) = \frac{\rho R^3}{3\varepsilon_0 x^2} \;(8)$$
+ while for smaller one, field is given by (4). Since $\vec{E_s}\uparrow\downarrow\vec{E_b}$, overlapping
+ $$E_{IV}(x,r') = E_b(x) - E_s(r') \;(9)$$
+ Putting (4) and (8) into (9)
+ $$E_{IV}(x,r') = \frac{\rho}{3\varepsilon_0}\left(\frac{R^3}{x^2}-\frac{r^3}{r'^2}\right)$$
+ but $l = x-r'$,
+ </p>
+ <h4>Answer 4</h4>
+ <p>
+ $$E_{IV}(x) = \frac{\rho}{3\varepsilon_0}\left[\frac{R^3}{x^2}-\frac{r^3}{(x-l)^2}\right]$$
+ for $x > R$.
+ </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
+ $$\vec{E}(x,r') = \vec{E_b}(x) + \vec{E_s}(r') = E_b(x) \hat{x} - E_s(r') \hat{r'} \;(10)$$
+ Substituting (1) and (4) into (10)
+ $$\vec{E}(x,r') = \frac{\rho}{3\varepsilon_0}(x\hat{x}-r'\hat{r'})$$
+ $$\vec{E}(x,r') = \frac{\rho}{3\varepsilon_0}(vec{x}-\vec{r'})$$
+ but $\vec{l} = \vec{x} -\vec{r'}$, hence
+ $$\vec{E} = \frac{\rho\vec{l}}{3\varepsilon_0}$$
+ </p>
+ <h4>Answer 5</h4>
+ <p>
+ Since E is constant, it is homogeneous inside cavity.
+ </p>
+
+ <p style="text-align: right; font-style: italic; font-size: 14;">
+ BSc. Luis Daniel Fernández Quintana<br>
+ Physics Department (FCNE)<br>
+ Universidad de Oriente, Cuba<br>
+ </p>
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