Flux through a hole in a sphere?

[brianaw26]

Flux through a hole in a sphere?

I have worked on this for a while and need some help. Answers or hints are appericated


A 6:1 uC charge located at the origin of a cartesian coordinate system is surrounded by a nonconducting hollow sphere of radius 9:1 cm. A drill with a radius of 1:02 mm is aligned along the z axis, and a hole is drilled
in the sphere. Calculate the electric flux through the hole.
Answer in units of N/m(sqd)=C.

 

[OlderDan]

Please read this

https://www.physicsforums.com/showthread.php?t=94379

Show what you have tried.

 

[kyle8921]

The electric flux through the hole in the sphere can be calculated using Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε0). In this case, the charge enclosed is the 6:1 uC charge at the origin.

To calculate the flux, we first need to determine the electric field at the surface of the sphere. This can be done using the equation E = Q/(4πε0r^2), where Q is the charge and r is the distance from the charge to the surface of the sphere. In this case, r = 9.1 cm + 1.02 mm = 9.202 cm.

Substituting the values into the equation, we get E = (6.1x10^-6 C)/(4π(8.854x10^-12 F/m)(0.09202 m)^2) = 2.29x10^6 N/C.

Now, to calculate the flux through the hole, we need to consider the area of the hole and the angle at which the electric field passes through it. Since the hole is drilled along the z axis, the electric field lines will be perpendicular to the surface of the hole. Thus, the angle between the electric field and the surface of the hole is 90 degrees.

The area of the hole can be calculated using the formula A = πr^2, where r is the radius of the hole. In this case, r = 1.02 mm = 0.00102 m, so the area of the hole is A = π(0.00102 m)^2 = 3.27x10^-6 m^2.

Finally, we can calculate the electric flux through the hole using the formula Φ = E*A*cos(θ), where θ is the angle between the electric field and the surface of the hole. In this case, θ = 90 degrees, so cos(θ) = 0. Thus, the electric flux is Φ = (2.29x10^6 N/C)(3.27x10^-6 m^2)(0) = 0 N/m^2.

Therefore, the electric flux through the hole in the sphere is 0 N/m^2 or 0 C, as there is no charge enclosed by the hole. This makes sense, as the hole is drilled