Is the magnitude of the electric field vector of a charged particle constant?

Question

From what I understand, electric field (in Newtons per Coulomb or Volts per Metre) is supposed to be an easier way to deal with the electric force between a charged object and a given point.

My question is, does the electric field vary as a function of distance from the charge, just as the force does according to Coulomb's Law?:

F = (q1*q2) / 4*Pi*(epsilon)*(r^2)

Where q1 and q2 are the charges, epsilon is the electric permittivity constant, and r is the distance.

Answer 1

Yup. Electric field is nothing more than,

F/q2 for the field of q1.

Answer 2

Yes, it varies. Suppose that the charge q1 creates the electric field, then: E = F/q2 so it is inverse proportionally to the distance from the charge q1. If distance changes from r1 to r2 = 2r1 then E2 = E1/4.

Answer 3

stick to Coulombs' electrostatic stress formulation: F = kqq'/r² the place, the electrostatic consistent: ok = a million/4??o = 9x10^9 N m²/C² We could compute the contribution of each and every of the expenses to the excellent stress: F1 = (9x10^9)(2*7x10^-12)/0.5² = 0.5 N (repulsive) F2 = (9x10^9)(-4*7x10^-12)/0.5² = -a million N (alluring) F(complete) = -0.5 N (alluring)