$$E_I(x) = \frac{\rho}{3\varepsilon_0}\left[x+\frac{r^3}{(l-x)^2}\right]$$ for $0\leq x\leq l-r$.

    <p>
       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)
        $$\vec{E}(x,r') = \frac{\rho}{3\varepsilon_0}(x\hat{x}-r'\hat{r'})$$ 
        $$\vec{E}(\vec{x},\vec{r'}) = \frac{\rho}{3\varepsilon_0}(\vec{x}-\vec{r'})$$
        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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