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MA_775_DIABLO [31]
4 years ago
8

A uniform charge density of 509 nC/m3 is distributed throughout a spherical volume of radius 6.03 cm. Consider a cubical Gaussia

n surface with its center at the center of the sphere. What is the electric flux through this cubical surface if its edge length is (a)4.00 cm and (b)16.8 cm?
Physics
1 answer:
Artemon [7]4 years ago
4 0

Explanation:

(a)  It is known that relation between charge and volume is as follows.

            q_{enclosed} = \rho V_{cube}

                       = (509 \times 10^{-9} C/m^{3}) \times (0.04 m)^{3}

                       = 509 \times 10^{-9} \times 6.4 \times 10^{-5}

                       = 3.26 \times 10^{-11} C

Now, according to Gauss's law

        \phi = \frac{q_{enclosed}}{\epsilon_{o}}

                   = \frac{3.26 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}

                   = 3.68 N m^{2}/C

Hence, the electric flux through this cubical surface if its edge length is 4.00 cm is 3.68 N m^{2}/C.

(b)   Similarly, we will calculate the electric flux when edge length is 16.8 cm as follows.

                q_{enclosed} = \rho V_{cube}

                       = (509 \times 10^{-9} C/m^{3}) \times (0.168 m)^{3}

                       = 509 \times 10^{-9} \times 4.74 \times 10^{-3}

                       = 2.41 \times 10^{-11} C

Now, according to Gauss's law

        \phi = \frac{q_{enclosed}}{\epsilon_{o}}

                   = \frac{2.41 \times 10^{-11} C}{8.85 \times 10^{-12}C^{2}N^{-1}m^{-2}}

                   = 2.72 N m^{2}/C

Therefore, the electric flux through this cubical surface if its edge length is 4.00 cm is 2.72 N m^{2}/C.

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