91影视

The given data is listed below,

• The charge on the inner wire is,$+Q$
• The charge on the outer wire is,$鈭�Q$
• The length of the wire is,$L=80\text{鈥塩尘}脳\frac{1\text{鈥尘}}{100\text{鈥塩尘}}=0.8\text{鈥尘}$
• The inner radius of the wire is,$r=0.7\text{鈥尘尘}脳\frac{1\text{鈥尘}}{1000\text{鈥尘尘}}=0.7脳{10}^{鈭�3}\text{鈥尘}$
• The outer radius of the wire is,$R=3\text{鈥塩尘}脳\frac{1\text{鈥尘}}{100\text{鈥塩尘}}=0.03\text{鈥尘}$

When two charged bodies collide, the charge flows from one conductor to the another. The electric potential is the electric state that governs the transfer of charge from one conductor to another that is in contact.

The potential difference should be negative because the field of the inner wire, which is approximately given by,

${\mathrm{E}}_{\mathrm{wire}}=\frac{2\mathrm{kQ}}{\mathrm{Lr}}$

Here, kis the coulomb constant with the value is$9脳{10}^9\text{鈥塏}鈰�{\text{m}}^2{\text{/C}}^2$ , Q is the charge on the wire, L is the length of the wire, and r is the inner radius of the wire.

The filed inside with tube is zero so it will not contribute in the potential difference so the final potential difference is given by,

$\begin{array}{c}螖V=V_{tube}鈭�V_{wire}\\ V_{tube}鈭�V_{wire}=鈭�\underset{r_1}{\overset{r_2}{鈭�}}Edr\\ =鈭�{鈭�}_r^R\frac{2kQ}{Lr}dr\\ =鈭�\frac{2kQ}{L}{鈭�}_r^R\frac{1}{r}dr\\ =鈭�\frac{2kQ}{L}\ln \left(\frac{R}{r}\right)\end{array}$

Thus, the potential difference${\mathrm{V}}_{\mathrm{tube}}鈭�{\mathrm{V}}_{\mathrm{wire}}$ is $鈭�\frac{2kQ}{L}\ln \left(\frac{R}{r}\right)$.

The electric charge on the wire is given by,

$\begin{array}{c}{\mathrm{E}}_{\mathrm{wire}}=\frac{2\mathrm{kQ}}{\mathrm{Lr}}\\ \mathrm{Q}=\frac{\mathrm{rLE}}{2\mathrm{k}}\end{array}$

Here, L is the length of the wire, E is the electric field for air breakdown in inner wire with an assumed value,$3脳{10}^6\text{鈥塏/颁}$ , and r is the inner radius of the wire.

Substitute all the values in the above equation.

$\begin{array}{c}Q=\frac{(0.7脳{10}^{鈭�3}\text{鈥尘})\left(0.8\text{鈥尘}\right)(3脳{10}^6\text{鈥塏/颁})}{2(9脳{10}^9\text{鈥塏}鈰�{\text{m}}^2{\text{/C}}^2)}\\ =9.35脳{10}^{鈭�8}\text{鈥塁}\end{array}$

The voltage supplied is given by,

$螖V=鈭�\frac{2kQ}{L}\ln \left(\frac{R}{r}\right)$

Substitute all the values in the above,

$\begin{array}{c}螖V=鈭�\frac{2(9脳{10}^9\text{鈥塏}鈰�{\text{m}}^2{\text{/C}}^2)(9.35脳{10}^{鈭�8}\text{鈥塁})}{\left(0.8\text{鈥尘}\right)}\ln \left(\frac{0.03\text{鈥尘}}{0.7脳{10}^{鈭�3}\text{鈥尘}}\right)\\ =鈭�7891.53\text{鈥塏}鈰�\text{m/C}\left(\frac{1\text{鈥塚}}{1\text{鈥塏}鈰�\text{m/C}}\right)\\ =鈭�7891.53\text{鈥塚}\end{array}$

Thus, the power supplied voltage is $鈭�7891.53\text{鈥塚}$.