187k views
1 vote
A 3.5-CM radius hemisphere contains a total charge of (6.6 times 10^-7) C. The flux through the rounded portion of the surface is (9.8 times 10^4 , textNcdottextm). The flux through the flat base is:

a) (9.8 times 10^4 , textNcdottextm)
b) (0 , textNcdottextm)
c) (4.9 times 10^4 , textNcdottextm)
d) (1.96 times 10^5 , textNcdottextm)

User Morgan Koh
by
7.6k points

1 Answer

3 votes

Final answer:

The total electric flux through a hemisphere is equal to the charge enclosed divided by the permittivity of free space. Since the flux through the rounded portion is provided, by symmetry, the flux through the flat base of the hemisphere is the same, which is (9.8 times 10^4 N·m^2/C).

Step-by-step explanation:

The question addresses the concept of electric flux through the surfaces of a hemisphere that encapsulates a total charge. According to Gauss's Law, the total electric flux Φ through a closed surface is equal to the charge Q enclosed by the surface divided by the permittivity of free space ε0:

Φ = Q / ε0

Since a hemisphere has two surfaces, the rounded dome and the flat base, the total electric flux through the entire hemispherical shape should be equal to the charge enclosed by it. Given that the flux through the rounded portion of the hemisphere is (9.8 times 10^4 N·m^2/C), and the total charge is the same throughout the hemisphere, by symmetry the flux through the flat base would be equal to the flux through the rounded dome. Therefore, the correct answer is:

a) (9.8 times 10^4 N·m^2/C)

User Krzysztof Bogdan
by
8.2k points