91影视

a) The radius of the solid cylinder, r=4.0 cm

b) Uniform surface charge density $蚁=A{r}^2$where,$A=2.5渭\mathrm{C}/{\mathrm{m}}^5$

Using the concept of volume charge density, we can get the total charge on the body. Then, using the concept of the electric flux theorem of the Gauss theorem, we can get the electric field at the required point. Now, for the second case, we needed to calculate the linear charge density of the material. Then, using this, we calculate the electric field at the point.

Formulae:

The volume of the solid cylinder,$V={\mathrm{蟺谤}}^2\mathrm{L}$ (1)

The electric flux of a conducting sheet, $蠒=\left|E\right|\left(2\mathrm{蟺谤L}\right)=\frac{{\mathrm{q}}_{\mathrm{enc}}}{{\mathrm{蔚}}_0}$ (2)

The electric field of a solid cylinder, $\left|E\right|=\frac{位}{2{\mathrm{蟺蔚}}_0\mathrm{r}}$ (3)

From equation (1), we can get the given equation as follows:

$dV=2\mathrm{蟺谤Ldr}$

The charge enclosed using the volume charge density and the above equation is given as:

$$\begin{gathered} q_{enc}={鈭�}_0^{\mathrm{r}}A{r}^{2}2\mathrm{蟺谤Ldr} \\ =\mathrm{A}2\mathrm{蟺尝}\underset{0}{\overset{\mathrm{r}}{鈭�}}{\mathrm{r}}^3\mathrm{dr} \\ =\frac{\mathrm{蟺}}{2}{\mathrm{ALr}}^4 \end{gathered}$$

Now, using the above equation in equation (ii), we get the electric field is given as:

$$\begin{gathered} \left|E\right|=\frac{A{r}^3}{4{蔚}_0} \\ =\frac{\left(2.5\mathrm{渭颁}/{\mathrm{m}}^5\right)脳\left(0.030\mathrm{m}\right)}{4脳\left(8.85脳{10}^{-12}\mathrm{F}/\mathrm{m}\right)} \\ =1.9\mathrm{N}/\mathrm{C} \end{gathered}$$

.

Hence, the value of the electric field is 1.9 N/C .

Using equation (a), the value of the linear charge density can be given as:

$$\begin{gathered} 位=\frac{1}{L}\underset{0}{\overset{0.04}{鈭�}}Ar2\mathrm{蟺尝dr} \\ =1.0脳{10}^{-11}\mathrm{C}/\mathrm{m} \end{gathered}$$

Now, the electric field at this point can be given using equation (3) as follows:

$$\begin{gathered} \left|\mathrm{E}\right|=\frac{2脳\left(9脳{10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\right)脳\left(1.0脳{10}^{-11}\mathrm{C}/\mathrm{m}\right)}{{\left(0.05\mathrm{m}\right)}^2} \\ =3.6\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is 3.6 N/C .