Answer:
1.![E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})](https://tex.z-dn.net/?f=E%28r%29%20%3D%20%5Cfrac%7B%5Calpha%7D%7B4%5Cpi%20%5Cepsilon_0%7D%282%20-%20%5Cfrac%7Br%7D%7BR%7D%29)
2.![E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}](https://tex.z-dn.net/?f=E%28r%29%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%5Cfrac%7B%5Calpha%20R%7D%7Br%7D)
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
![\rho (r) = \alpha (1-\frac{r}{R})](https://tex.z-dn.net/?f=%5Crho%20%28r%29%20%3D%20%5Calpha%20%281-%5Cfrac%7Br%7D%7BR%7D%29)
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
![\int {\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}](https://tex.z-dn.net/?f=%5Cint%20%7B%5Cvec%7BE%7D%7D%20%5C%2C%20d%5Cvec%7Ba%7D%20%3D%20%5Cfrac%7BQ_%7Benc%7D%7D%7B%5Cepsilon_0%7D)
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
![\int\, da = 2\pi r h](https://tex.z-dn.net/?f=%5Cint%5C%2C%20da%20%3D%202%5Cpi%20r%20h)
where ‘h’ is the length of the imaginary Gaussian surface.
![Q_{enc} = \int\limits^r_0 {\rho(r)h} \, dr = \alpha h \int\limits^r_0 {(1-r/R)} \, dr = \alpha h (r - \frac{r^2}{2R})\left \{ {{r=r} \atop {r=0}} \right. = \alpha h (\frac{2Rr - r^2}{2R})\\E2\pi rh = \alpha h \frac{2Rr - r^2}{2R\epsilon_0}\\E(r) = \alpha \frac{2R - r}{4\pi \epsilon_0 R}\\E(r) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R})](https://tex.z-dn.net/?f=Q_%7Benc%7D%20%3D%20%5Cint%5Climits%5Er_0%20%7B%5Crho%28r%29h%7D%20%5C%2C%20dr%20%3D%20%5Calpha%20h%20%5Cint%5Climits%5Er_0%20%7B%281-r%2FR%29%7D%20%5C%2C%20dr%20%3D%20%5Calpha%20h%20%28r%20-%20%5Cfrac%7Br%5E2%7D%7B2R%7D%29%5Cleft%20%5C%7B%20%7B%7Br%3Dr%7D%20%5Catop%20%7Br%3D0%7D%7D%20%5Cright.%20%3D%20%5Calpha%20h%20%28%5Cfrac%7B2Rr%20-%20r%5E2%7D%7B2R%7D%29%5C%5CE2%5Cpi%20rh%20%3D%20%5Calpha%20h%20%5Cfrac%7B2Rr%20-%20r%5E2%7D%7B2R%5Cepsilon_0%7D%5C%5CE%28r%29%20%3D%20%5Calpha%20%5Cfrac%7B2R%20-%20r%7D%7B4%5Cpi%20%5Cepsilon_0%20R%7D%5C%5CE%28r%29%20%3D%20%5Cfrac%7B%5Calpha%7D%7B4%5Cpi%20%5Cepsilon_0%7D%282%20-%20%5Cfrac%7Br%7D%7BR%7D%29)
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
![Q_{enc} = \int\limits^R_0 {\rho(r)h} \, dr = \alpha \int\limits^R_0 {(1-r/R)h} \, dr = \alpha h(r - \frac{r^2}{2R})\left \{ {{r=R} \atop {r=0}} \right. = \alpha h(R - \frac{R^2}{2R}) = \alpha h\frac{R}{2} \\E2\pi rh = \frac{\alpha Rh}{2\epsilon_0}\\E(r) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r}](https://tex.z-dn.net/?f=Q_%7Benc%7D%20%3D%20%5Cint%5Climits%5ER_0%20%7B%5Crho%28r%29h%7D%20%5C%2C%20dr%20%3D%20%5Calpha%20%5Cint%5Climits%5ER_0%20%7B%281-r%2FR%29h%7D%20%5C%2C%20dr%20%3D%20%5Calpha%20h%28r%20-%20%5Cfrac%7Br%5E2%7D%7B2R%7D%29%5Cleft%20%5C%7B%20%7B%7Br%3DR%7D%20%5Catop%20%7Br%3D0%7D%7D%20%5Cright.%20%3D%20%5Calpha%20h%28R%20-%20%5Cfrac%7BR%5E2%7D%7B2R%7D%29%20%3D%20%5Calpha%20h%5Cfrac%7BR%7D%7B2%7D%20%5C%5CE2%5Cpi%20rh%20%3D%20%5Cfrac%7B%5Calpha%20Rh%7D%7B2%5Cepsilon_0%7D%5C%5CE%28r%29%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%5Cfrac%7B%5Calpha%20R%7D%7Br%7D)
3. At the boundary where r = R:
![E(r=R) = \frac{\alpha}{4\pi \epsilon_0}(2 - \frac{r}{R}) = \frac{\alpha}{4\pi \epsilon_0}\\E(r=R) = \frac{1}{4\pi \epsilon_0}\frac{\alpha R}{r} = \frac{\alpha}{4\pi \epsilon_0}](https://tex.z-dn.net/?f=E%28r%3DR%29%20%3D%20%5Cfrac%7B%5Calpha%7D%7B4%5Cpi%20%5Cepsilon_0%7D%282%20-%20%5Cfrac%7Br%7D%7BR%7D%29%20%3D%20%5Cfrac%7B%5Calpha%7D%7B4%5Cpi%20%5Cepsilon_0%7D%5C%5CE%28r%3DR%29%20%3D%20%5Cfrac%7B1%7D%7B4%5Cpi%20%5Cepsilon_0%7D%5Cfrac%7B%5Calpha%20R%7D%7Br%7D%20%3D%20%5Cfrac%7B%5Calpha%7D%7B4%5Cpi%20%5Cepsilon_0%7D)
As can be seen from above, two E-field values are equal as predicted.