Answer :
Answer:
a) [tex]\sigma_{\rm in} = -2.18~{\rm \mu C/m^2}[/tex]
b) [tex]\sigma_{\rm out}= 1.12~{\rm \mu C/m^2}[/tex]
c) [tex]E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}[/tex]
Explanation:
Before the wire is inserted, the total charge on the inner and outer surface of the cylindrical shell is as follows:
[tex]Q_{\rm in} = \sigma A_{\rm in} = \sigma(2\pi r_1 h) = (-0.35)(2\pi (0.08) h) = -0.175h~{\rm \mu C}[/tex]
[tex]Q_{\rm out} = \sigma A_{\rm out} = \sigma(2\pi r_2 h) = (-0.35)(2\pi (0.1) h) = -0.22h~{\rm \mu C}[/tex]
Here, 'h' denotes the length of the cylinder. The total charge of the cylindrical shell is -0.395h μC.
When the thin wire is inserted, the positive charge of the wire attracts the same amount of negative charge on the inner surface of the shell.
[tex]Q_{\rm wire} = \lambda h = 1.1h~{\rm \mu C}[/tex]
a) The new charge on the inner shell is -1.1h μC. Therefore, the new surface charge density of the inner shell can be calculated as follows:
[tex]\sigma_2 = \frac{Q_{\rm in}}{2\pi r_1h} = \frac{-1.1h}{2\pi r_1 h} = \frac{-1.1}{2\pi(0.08)} = -2.18~{\rm \mu C/m^2}[/tex]
b) The new charge on the outer shell is equal to the total charge minus the inner charge. Therefore, the new charge on the outer shell is +0.705 μC.
The new surface charge density can be calculated as follows:
[tex]\sigma_{\rm out}= \frac{Q_{\rm out}}{2\pi r_2h} = \frac{0.705h}{2\pi r_2 h} = \frac{0.705}{2\pi(0.1)} = 1.12~{\rm \mu C/m^2}[/tex]
c) The electric field outside the cylinder can be found by Gauss' Law:
[tex]\int{\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}[/tex]
We will draw an imaginary cylindrical shell with radius r > r2. The integral in the left-hand side will be equal to the area of the imaginary surface multiplied by the E-field.
[tex]E(2\pi r h) = \frac{Q_{\rm enc}}{\epsilon_0}\\E2\pi rh = \frac{\sigma 2\pi (r_1 + r_2)h}{\epsilon_0}\\E = \frac{\sigma(r_1 + r_2)}{\epsilon_0 r}[/tex]