Apoint charge q +4 μc is at the center of a sphere of radius 0.4 m. (a) find the surface area of the sphere. 2.01 (b) find the magnitude of the electric field at points on the surface of the sphere. m2 x n/c (c) what is the flux of the electric field due to the point charge through the surface of the sphere? n -m2/c d) would your answer to part (c) change if the polnt charge were moved so that it was inslde the sphere but not at its center? yes no not enough information to decide (e) what is the net flux through a cube of side 2 m that encloses the sphere?

area  A = 2.01 m² ,  electric fields E = 2.25 10⁵ N/C and flux   Φ = 4.52 10⁵ N m²/C

Explanation:

The electric field of a point charge can be calculated with the equation

       E = k q / r²

Where q is the point charge, r is the distance of the charge to the sphere and k the constant Coulomb is  8.99 10⁹ N m² / C²

a) The surface or area of a sphere is

       A = 4π R²

       A = 4 π 0.4²

       A = 2.01 m²

b) We use the electric field equation

      E = 8.99 10⁹ 4 10⁻⁶ / 0.4²

      E = 2.25 10⁵ N / C

c) The electric flow can be calculated by the equation

     Φ = ∫ E. dA

The point is the scalar product, the electric field lines leave the charge and as the charge is in the center of the sphere, it co-exists with the radii of this, which are the lines of the area, so they are parallel and the scalar product it is reduced to the ordinary product

      Φ = E A

      Φ = 2.25 10⁵ 2.01

      Φ = 4.52 10⁵ N m²/C

d) If we write Gauss's law

       Φ = E A = Qint / ε₀

When analyzing this law the flow is proportional to the value of the charge inside the spheres, regardless of the point where this face is located, so if the charge moves a little it will not change the total flow on the sphere.

e) Let's use Gauss's law for this part

         Φ = Qint / eo

Regardless of the shape of the surface

        Φ = 4 10⁻⁶ / 8.85 10⁻¹²  

        Φ = 4.52 10⁵ N m² / C

nah.   i'm just sleepy.

a=d/t

a="triangle"v/t

explanation:

a = acceleration