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1. What is the electric flux passing through a Gaussian surface that contains 2 protons and 1 electron?
2. A solid conducting sphere with an area of 16 m² has a total charge of 4 µC. Determine the electric field through the Gaussian surface.
3. Solve for the distant planet, Uhkbarhr, has a net electric flux of -4.65 X 10¹⁹ Nm²/C. What will be the total electric charge on the planet?
4. A copper wire that has a radius of 3 cm has a uniform charge density of 12 µC/m². Determine the total charge present in a 1.0-m long section of the copper wire.
5. Solve the magnitude of the electric field in the previous problem at a distance of 12.13 cm from the wire.
1. What is the electric flux passing through a Gaussian surface that contains 2 protons and 1 electron?
2. A solid conducting sphere with an area of 16 m² has a total charge of 4 µC. Determine the electric field through the Gaussian surface.
3. Solve for the distant planet, Uhkbarhr, has a net electric flux of -4.65 X 10¹⁹ Nm²/C. What will be the total electric charge on the planet?
4. A copper wire that has a radius of 3 cm has a uniform charge density of 12 µC/m². Determine the total charge present in a 1.0-m long section of the copper wire.
5. Solve the magnitude of the electric field in the previous problem at a distance of 12.13 cm from the wire.
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The electric flux passing through a Gaussian surface depends on the net charge enclosed within the surface. Since the Gaussian surface contains 2 protons and 1 electron, the net charge enclosed is +2e - (-1e) = +3e, where e is the elementary charge. Therefore, the electric flux passing through the Gaussian surface depends on the shape of the surface and the net charge enclosed, and cannot be determined with the given information.
The electric field through a Gaussian surface depends on the total charge enclosed within the surface and the shape of the surface. In this case, the solid conducting sphere has a total charge of 4 µC. Since the sphere is a conductor, the charge is uniformly distributed over its surface, and the electric field inside the sphere is zero. Therefore, the electric field through the Gaussian surface is also zero.
The net electric flux passing through a closed surface is proportional to the net charge enclosed within the surface, according to Gauss's law. Therefore, if the net electric flux on the planet Uhkbarhr is -4.65 X 10¹⁹ Nm²/C, then the net charge enclosed within the closed surface is also -4.65 X 10¹⁹ C.
The total charge Q present in a 1.0-m long section of the copper wire can be calculated as follows:
Determine the area A of the cross-section of the wire: A = πr² = π(0.03 m)² ≈ 0.00283 m².
Determine the total charge q on the cross-section of the wire: q = ρAℓ, where ρ is the charge density, A is the area, and ℓ is the length of the wire. Substituting ρ = 12 µC/m², A = 0.00283 m², and ℓ = 1.0 m, we get q = (12 µC/m²)(0.00283 m²)(1.0 m) = 0.034 µC.
Therefore, the total charge present in the 1.0-m long section of the copper wire is Q = |q| = 0.034 µC.
The electric field E at a distance r from an infinitely long, uniformly charged wire of linear charge density λ is given by E = λ/(2πε₀r), where ε₀ is the permittivity of free space. In this case, the wire has a charge density of 12 µC/m², which is equivalent to a linear charge density of λ = 0.012 C/m. Substituting this value, along with r = 12.13 cm = 0.1213 m, and the value of ε₀, we get:
E = (0.012 C/m)/(2π(8.85 x 10^-12 N^-1 m^-2)(0.1213 m)) ≈ 45.5 N/C
Therefore, the magnitude of the electric field at a distance of 12.13 cm from the wire is approximately 45.5 N/C