A hemispherical surface (half of a spherical surface) of radius (𝑅=30cm) is located in a uniform electric field ofmagnitude (𝐸=20 N/m) that is parallel to the axis of symmetry of the hemisphere. What is the magnitude of theelectric flux through the hemispherical surface?Select one:a. 11.3b. 7.54c. none of the answers d. 5.65e. 3.77

The electric flux through a surface is given by the equation Φ = E * A * cos(θ), where E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface.

In this case, the electric field E is given as 20 N/m. The area A of a hemisphere is 2πR^2, and for a hemisphere of radius R = 30 cm = 0.3 m, the area is A = 2π(0.3 m)^2 = 0.565 m^2. The angle θ between the electric field and the normal to the surface is 0 degrees for a hemisphere, so cos(θ) = 1.

Substituting these values into the equation for electric flux gives Φ = 20 N/m * 0.565 m^2 * 1 = 11.3 N*m^2/C.

So, the correct answer is a. 11.3.

The electric flux through a surface is given by the equation Φ = E * A * cos(θ), where E is the electric field, A is the area of the surface, and θ is the angle between the electric field and the normal to the surface.

In this case, the electric field E is given as 20 N/m. The area A of a hemisphere is 2πR^2, where R is the radius of the hemisphere. Given that R = 30 cm = 0.3 m, we can calculate A as:

A = 2π(0.3 m)^2 = 0.565 m^2.

The angle θ between the electric field and the normal to the surface is 0 degrees, because the electric field is parallel to the axis of symmetry of the hemisphere. Therefore, cos(θ) = cos(0) = 1.

Substituting these values into the equation for electric flux gives:

Φ = E * A * cos(θ) = 20 N/m * 0.565 m^2 * 1 = 11.3 N*m^2/C.

Therefore, the magnitude of the electric flux through the hemispherical surface is 11.3 N*m^2/C. So, the correct answer is (a) 11.3.