Consider a point charge qqq in three-dimensional space. Symmetry requires the electric field to point directly away from the charge in all directions. To find E(r)E(r)E(r), the magnitude of the field at distance rrr from the charge, the logical Gaussian surface is a sphere centered at the charge. The electric field is normal to this surface, so the dot product of the electric field and an infinitesimal surface element involves cos(0)=1cos⁡(0)=1. The flux integral is therefore reduced to ∫E(r)dA=E(r)A(r)∫E(r)dA=E(r)A(r), where E(r)E(r)E(r) is the magnitude of the electric field on the Gaussian surface, and A(r)A(r)A(r) is the area of the surface.

 E =

Explanation:

For this exercise let's use Gauss's law. The Gaussian surface that follows the symmetry of the charges is a sphere

           Ф = ∫ E. dA =

the bold are vectors, the radii of the sphere and the electric field are parallel therefore the scalar product reduces to the algebraic product

           Ф = ∫ E dA = \frac{x_{int} }{\epsilon_o}

           E ∫ dA = \frac{x_{int} }{\epsilon_o}

           E A = \frac{x_{int} }{\epsilon_o}

the area of a sphere is

            A = 4π r²

the charge inside the sphere is q = + q

we substitute

           E 4π r² = \frac{x }{\epsilon_o}

           E =

true, as the earth rotates the sun hits the surface of the planet; causing the day and night cycle.

50 inches should be the resultant vector