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Answer :
The surface charge density on the outer shell is approximately [tex]\( 9.09 \times 10^{-6} \, \text{C/m}^2 \)[/tex], and the surface charge density on the inner shell is approximately [tex]\( 8.23 \times 10^{-6} \, \text{C/m}^2 \).[/tex]
To solve this problem, we can use the formula for the capacitance of a spherical capacitor:
[tex]\[ C = \frac{4\pi\epsilon_0}{\frac{1}{r_1} - \frac{1}{r_2}} \][/tex]
Where:
- [tex]\( C \)[/tex] is the capacitance
- [tex]\( \epsilon_0 \)[/tex] is the permittivity of vacuum [tex](\( \epsilon_0 \approx 8.85 \times 10^{-12} \, \text{F/m} \))[/tex]
- [tex]\( r_1 \)[/tex] is the inner radius of the outer shell
- [tex]\( r_2 \)[/tex] is the outer radius of the outer shell
Given that the capacitance [tex]\( C = 125 \, \text{pF} = 125 \times 10^{-12} \, \text{F} \)[/tex] and the inner radius [tex]\( r_1 = 9.00 \, \text{cm} = 0.09 \, \text{m} \)[/tex], we can rearrange the formula to solve for [tex]\( r_2 \):[/tex]
[tex]\[ r_2 = \frac{1}{\frac{1}{r_1} - \frac{C}{4\pi\epsilon_0}} \][/tex]
Now, we can calculate the value of [tex]\( r_2 \)[/tex]:
[tex]\[ r_2 = \frac{1}{\frac{1}{0.09} - \frac{125 \times 10^{-12}}{4\pi \times 8.85 \times 10^{-12}}} \][/tex]
[tex]\[ r_2 \approx 0.1213 \, \text{m} \][/tex]
Now that we have both radii, we can use the formula for the surface charge density [tex]\( \sigma \)[/tex]on a spherical conductor:
[tex]\[ \sigma = \frac{Q}{A} \][/tex]
Where:
- [tex]\( Q \)[/tex] is the charge
- [tex]\( A \)[/tex] is the surface area of the sphere
a. To find the surface charge density on the outer shell, we need to find the charge [tex]\( Q \)[/tex] on the outer shell. Since the potential difference between the shells is given and the capacitance is known, we can use the formula for the capacitance:
[tex]\[ C = \frac{Q}{V} \][/tex]
Where [tex]\( V \)[/tex] is the potential difference between the shells.
[tex]\[ Q = C \times V \][/tex]
[tex]\[ Q = (125 \times 10^{-12}) \times 355 \][/tex]
[tex]\[ Q = 4.4375 \times 10^{-8} \, \text{C} \][/tex]
Now, we can find the surface charge density on the outer shell:
[tex]\[ \sigma = \frac{Q}{4\pi r_2^2} \][/tex]
[tex]\[ \sigma = \frac{4.4375 \times 10^{-8}}{4\pi (0.1213)^2} \][/tex]
[tex]\[ \sigma \approx 9.09 \times 10^{-6} \, \text{C/m}^2 \][/tex]
b. To find the surface charge density on the inner shell, we use the same charge [tex]\( Q \)[/tex] as calculated above. The surface area [tex]\( A \)[/tex] of the inner shell is [tex]\( 4\pi r_1^2 \).[/tex]
[tex]\[ \sigma = \frac{Q}{4\pi r_1^2} \][/tex]
[tex]\[ \sigma = \frac{4.4375 \times 10^{-8}}{4\pi (0.09)^2} \][/tex]
[tex]\[ \sigma \approx 8.23 \times 10^{-6} \, \text{C/m}^2 \][/tex]
Therefore, the surface charge density on the outer shell is approximately [tex]\( 9.09 \times 10^{-6} \, \text{C/m}^2 \)[/tex], and the surface charge density on the inner shell is approximately [tex]\( 8.23 \times 10^{-6} \, \text{C/m}^2 \).[/tex]
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