A 14 cm-radius hemisphere contains a total charge of853.1 10 C− . The flux through therounded portion of the surface is4 211 10 /Nm C . What is the flux through the flat baseof the hemisphere?
Question
A 14 cm-radius hemisphere contains a total charge of853.1 10 C− . The flux through therounded portion of the surface is4 211 10 /Nm C . What is the flux through the flat baseof the hemisphere?
Solution
The total electric flux through a closed surface containing a charge Q is given by Gauss's law as Φ = Q/ε0.
Given that the total charge Q in the hemisphere is 853.1 x 10^-6 C and the flux through the rounded portion of the hemisphere is 211 x 10^4 Nm^2/C, we can find the total flux through the entire hemisphere and then subtract the flux through the rounded portion to find the flux through the flat base.
First, we need to know the value of ε0, the permittivity of free space. This is a constant with a value of approximately 8.85 x 10^-12 C^2/Nm^2.
Using Gauss's law, the total flux Φ_total through the hemisphere is:
Φ_total = Q/ε0 = (853.1 x 10^-6 C) / (8.85 x 10^-12 C^2/Nm^2) = 9.64 x 10^5 Nm^2/C
The flux through the flat base of the hemisphere Φ_base is then:
Φ_base = Φ_total - Φ_rounded = 9.64 x 10^5 Nm^2/C - 211 x 10^4 Nm^2/C = 7.53 x 10^5 Nm^2/C
So, the flux through the flat base of the hemisphere is 7.53 x 10^5 Nm^2/C.
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