A non-conducting cube of side length 1m has a spherical cavity of radius 0.1m inside it. The system has a charge density of 2.5 C/m3. What is the total flux coming out of the surface of the cube?
Question
A non-conducting cube of side length 1m has a spherical cavity of radius 0.1m inside it. The system has a charge density of 2.5 C/m3. What is the total flux coming out of the surface of the cube?
Solution
The total electric flux Φ through a closed surface in an electric field is given by Gauss's law:
Φ = Q/ε₀
where Q is the total charge enclosed by the surface and ε₀ is the permittivity of free space.
First, we need to calculate the total charge in the cube. The charge density ρ is given as 2.5 C/m³. The volume V of the cube is its side length cubed, which is 1m³. So, the total charge Q in the cube is ρV = 2.5 C/m³ * 1 m³ = 2.5 C.
However, we need to subtract the charge that would have been in the spherical cavity, because there is no charge in the cavity. The volume of the cavity is given by the formula for the volume of a sphere, 4/3πr³. The radius r of the cavity is 0.1m, so the volume of the cavity is 4/3π(0.1m)³ = 0.00419 m³. The charge that would have been in the cavity is therefore ρV = 2.5 C/m³ * 0.00419 m³ = 0.01048 C.
So, the total charge Q enclosed by the surface of the cube is 2.5 C - 0.01048 C = 2.48952 C.
Finally, we can calculate the total flux Φ. The permittivity of free space ε₀ is approximately 8.85 x 10^-12 C²/Nm². So, the total flux Φ is Q/ε₀ = 2.48952 C / 8.85 x 10^-12 C²/Nm² = 2.81 x 10^11 Nm²/C.
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