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IgorLugansk [536]
3 years ago
15

A conducting spherical shell having an inner radius of 3 cm and outer radius of 4.3 cm carries a net charge of 9.6 μC. A conduct

ing sphere is placed at the center of this shell hav- ing a radius of 0.8 cm and carries a net charge of 1.7 μC. Determine the surface charge density on the outer surface of the shell.
Engineering
1 answer:
Arte-miy333 [17]3 years ago
4 0

Answer:

sigma = 3.4*10^-4 C / m^2

Explanation:

Given:

- The net charge of the conducting sphere Q_s = 1.7 uC

- The net charge in the shell Q_shell = 9.6 uC

- Outer radius of the shell r = 4.3 cm

Find:

Determine the surface charge density on the outer surface of the shell.

Solution:

- The concept here is that the net charge on the conducting sphere must be equal to charge on inner shell. Hence, Q_s = Q_in

                                   Q_shell = Q_in + Q_out

                                   Q_out = Q_shell - Q_in

                                   Q_out = 9.6 - 1.7 = 7.9 uC

- Now we can calculate the charge density of the outer shell by:

                                   sigma = Q_out / A_out

                                   sigma = (7.9*10^-6) / 4*pi*(0.043)^2

                                  sigma = 3.4*10^-4 C / m^2

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It has been estimated that 139.2x10^6 m^2 of rainforest is destroyed each day. assume that the initial area of tropical rainfore
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Answer:

A. 6.96 x 10^-6 /day

B. 22.466 x 10^12 m^2

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Therefore to calculate exponential rate in 1/day,

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= 6.96 x 10^-6 /day

B. Rainforest left in 2015 using the rate in A.

2015 - 1975 = 40 years

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= 14610 days

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Rate of rainforest destruction = 139.2 x 10^6 m^2/day

Area of rainforest in 2015 = 14610 * 139.2 x 10^6

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Area left = area of rainforest in 1975 - area of rainforest destroyed in 40years

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So CO2 removed is the same as rainforest removed so we use the rate of rainforest removed in a day

Area of rainforest in 1975 = 24.5 x 10^12 m^2

Area of rainforest removed in 2025 = 18262 days * 139.2 x 10^6

= 2.54 x 10^12 m^2

Area of rainforest removed between 1975 - 2025 = 24.5 x 10^12 - 2.54 x 10^12

= 21.958 x 10^12 mC2 of rainforest removed

CO2 = 0.83kg/m^2.year

CO2 removed between 1975 - 2025 = 0.83 * 21.958 x 10^12 * 50 years

= 9.1125 x 10^14 kg of CO2 was removed between 1975 - 2025

6 0
3 years ago
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