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ser-zykov [4K]
4 years ago
5

The electric field everywhere on the surface of a thin, spherical shell of radius 0.710 m is of magnitude 936 N/C and points rad

ially toward the center of the sphere. (a) What is the net charge within the sphere's surface? nC (b) What is the distribution of the charge inside the spherical shell? The positive charge has an asymmetric charge distribution. The negative charge has an asymmetric charge distribution. The positive charge has a spherically symmetric charge distribution. The negative charge has a spherically symmetric charge distribution.
Physics
1 answer:
kaheart [24]4 years ago
5 0

Explanation:

Given that,

Radius of a spherical shell, r = 0.71 m

Electric field that points radially toward the center of the sphere is 936 N/C

(a) Let q is the net charge within the sphere's surface. Using Gauss's law, we can find it :

\dfrac{1}{q}\times \epsilon_0=EA\\\\q=\dfrac{\epsilon_0}{EA}\\\\q=\dfrac{8.85\times 10^{-12}}{936\times \pi \times (0.71)^2}\\\\q=5.97\times 10^{-15}\ C

(b) Gauss's law is used to find the amount of charge enclosed within a surface. It doesn't say anything about the distribution of the charge inside the spherical shell.

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