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A hollow conducting sphere with an outer radius of 0.295 m and an inner radius of 0.200 m has a uniform surface charge density o

IrinaK [193]

Answer:

a. $6.032\times10^{-6}C/m^2$

b. $6.816\times10^5N/C$

Explanation:

#Apply surface charge density, electric field, and Gauss law to solve:

a. Surface charge density is defined as charge per area denoted as $\sigma$

$\sigma=\frac{Q}{4\pi r_{out}^2}$ , and the strength of the electric field outside the sphere $E=\frac{\sigma _{new}}{\epsilon _o}$

Using Gauss Law, total electric flux out of a closed surface is equal to the total charge enclosed divided by the permittivity.

$\phi=\frac{Q_{enclosed}}{\epsilon_o}\\\\\sigma=\frac{Q}{4\pi r_{out}^2}\\\\\sigma=\frac{0.370\times 10^{-6}}{4\pi \times (0.295m)^2}\\\\=3.383\times10^{-7}C/m^2$ #surface charge outside sphere.

$\sigma_{new}=\sigma_{s}-\sigma\\\\\sigma_{new}=6.37\times10^{-6}C/m^2-3.383\times10^{-7}C/m^2\\\\\sigma_{new}=6.032\times10^{-6}C/m^2$

Hence, the new charge density on the outside of the sphere is $6.032\times10^{-6}C/m^2$

b. The strength of the electric field just outside the sphere is calculated as:

From a above, we know the new surface charge to be $6.032\times10^{-6}C/m^2$ ,

$E=\frac{\sigma _{new}}{\epsilon _o}\\\\=\frac{6.032\times10^{-6}C/m^2}{\epsilon _o}\\\\\epsilon _o=8.85\times10^{-12}C^2/N.m^2\\\\E=\frac{6.032\times10^{-6}C/m^2}{8.85\times10^{-12}C^2/N.m^2}\\\\E=6.816\times10^5N/C$

Hence, the strength of the electric field just outside the sphere is $6.816\times10^5N/C$

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lesya692 [45]

Answer:

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Explanation:

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