Hen a test charge of + 5.00 nC is placed at a certain point, the force that acts on it has a magnitude of 0.0800 N and is directed northeast. (a) If the test charge were −2.00 nC instead, what force would act on it? (b) What is the electric field at the point in question?

a) F = 0.032 N, The direction is opposite to the field,  45 south west

b) E  = 1.6 10⁷ N / c, With northeast direction

Explanation:

The electric force is given by the expression

           F = q E

Let's look for the elective field with the initial data

           E = F / q

Let's reduce the magnitudes to the SI system

          q = 5.00 nC = 5.00 10⁻⁹ C

Let's calculate

        E = 0.0800 / 5.00 10⁻⁹

        E = 0.016 10⁹N / C = 1.6 10⁷ N / c

With northeast direction

a) If we change the test charge for another of

        q₂ = -2 nC = -2.0 10⁻⁹ C

        F = -2.0 10⁻⁹  1.6 10⁷

        F = 3.2 10⁻² N = 0.032 N

The direction is opposite to the field, bone 45 south west

compass can be used day and night, cloudy or sunny