A point charge causes an electric flux of3 2 12 10 Nm C− to pass through a sphericalGaussian surface of 20.0 cm radius centered on the charge. If the radius of the Gaussiansurface were doubled, how much flux would pass through the surface?
Question
A point charge causes an electric flux of3 2 12 10 Nm C− to pass through a sphericalGaussian surface of 20.0 cm radius centered on the charge. If the radius of the Gaussiansurface were doubled, how much flux would pass through the surface?
Solution
The electric flux through a Gaussian surface is given by Gauss's Law, which states that the electric flux through any closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space (ε₀). Mathematically, this is expressed as:
Φ = Q / ε₀
In this case, the charge Q causing the flux can be found from the given flux Φ and the known value of ε₀ (approximately 8.85 x 10^-12 C^2/Nm^2) as:
Q = Φ * ε₀
Doubling the radius of the Gaussian surface does not change the amount of charge enclosed by the surface, so the total electric flux through the surface remains the same. Therefore, if the radius of the Gaussian surface were doubled, the flux passing through the surface would still be 3.2 x 10^12 Nm/C.
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