91影视

a) The radius of the conducting rod ${\mathrm{R}}_1=1.30\mathrm{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=1.30脳{10}^{-3}\mathrm{m}$,

b) Length of the rod, L=11.00 m

c) The net charge on the rod, role="math" localid="1657342970971" $Q_1=+3.40脳{10}^{-12}C$

d) The net charge on the shell,$Q_2=-6.80脳{10}^{-12}C$

Using the concept of the Gauss flux theorem, we can get the electric field passing through the enclosed surface considering the enclosed charge in the surface. Now, using the value of the electric field, we can get the inner and outer charge of the shell.

Formula:

The electric charge enclosed within the Gaussian surface, $q_{enc}={蔚}_0蠒=2{\mathrm{蟺蔚}}_0\mathrm{LE}$ (1)

We take the Gaussian surface to be a cylinder of length L, coaxial with the given cylinders, and of radius r. We may ignore any flux through the ends. In this case, we take the radius of our Gaussian cylinder to be:

$$\begin{gathered} \mathrm{r}=2.00{\mathrm{R}}_2 \\ =\left(2.00\right)\left(10.0{\mathrm{R}}_1\right) \\ =\left(2.00\right)\left(10.0脳1.3脳{10}^{-3}\mathrm{m}\right) \\ =2.6脳{10}^{-2}\mathrm{m} \end{gathered}$$

The charge enclosed in this case is given as:

$$\begin{gathered} {\mathrm{q}}_{\mathrm{enc}}=-{\mathrm{Q}}_1 \\ =-3.40脳{10}^{-12}\mathrm{C} \end{gathered}$$

Now, using equation (1), we can get the electric field for this case as given:

role="math" localid="1657343754225" $$\begin{gathered} \mathrm{E}=\frac{-3.40脳{10}^{-12}\mathrm{C}}{2\mathrm{蟺}\left(8.85脳{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\left(11.0\mathrm{m}\right)\left(2.6脳{10}^{-2}\mathrm{m}\right)} \\ =-0.214\mathrm{N}/\mathrm{C} \end{gathered}$$.

Hence, the magnitude of the electric field is 0.214 N/C .

From the calculations of part (a), the negative sign in the electric field indicates that the field points inward.

In this case, we take the radius of our Gaussian cylinder to be:

$$\begin{gathered} \mathrm{r}=5.00{\mathrm{R}}_1 \\ =5.00脳\left(1.3脳{10}^{-3}\mathrm{m}\right) \\ =6.5脳{10}^{-3}\mathrm{m} \end{gathered}$$

The charge enclosed by the Gaussian surface is given as:

$$\begin{gathered} q_{enc}=Q_1 \\ =+3.40脳{10}^{-12}{C}^{\text{'}} \end{gathered}$$

The electric field in this case using equation (1) is given as:

$$\begin{gathered} \mathrm{E}=\frac{-3.40脳{10}^{-12}\mathrm{C}}{2\mathrm{蟺}\left(8.85脳{10}^{-12}{\mathrm{C}}^2/\mathrm{N}.{\mathrm{m}}^2\right)\left(11.0\mathrm{m}\right)\left(6.50脳{10}^{-3}\mathrm{m}\right)} \\ =-0.855\mathrm{N}/\mathrm{C} \end{gathered}$$

Hence, the value of the electric field is 0.855N/C .

From the calculation of part (c), the positive sign of the electric field indicates that the field points outward.

We consider a cylindrical Gaussian surface whose radius places it within the shell itself. The electric field is zero at all points on the surface since any field within a conducting material would lead to current flow (and thus to a situation other than the electrostatic ones being considered here), so the total electric flux through the Gaussian surface is

zero and the net charge within it is zero (by Gauss鈥� law). Since the central rod has a charge $Q_1$, the inner surface of the shell must have a charge value as:

$$\begin{gathered} Q_{in}=-Q_1 \\ =-3.40脳{10}^{-12}C \end{gathered}$$

Hence, the value of the charge is $-3.40脳{10}^{-12}C$.

Since the shell is known to have a total charge as: $Q_2=-2.00Q_1$

Thus, it must have a charge on the outer shell as given:

role="math" localid="1657344081621" $$\begin{gathered} Q_{out}=Q_2-Q_{in} \\ =-Q_1 \\ =-3.40脳{10}^{-12}C \end{gathered}$$

.

Hence, the value of the charge is $-3.40脳{10}^{-12}C$.