Question

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?

Answer 1

a) -0.03 N b) 1.6*10⁷ N/C

Step-by-step explanation:

a) Assuming that the test charge doesn't perturb the charge distribution that creates the force, this force can be expressed as the product of the charge times the electric field at this point:

F = q*E = 0.0800 N

⇒
`E = (F)/(q) =(0.0800N)/(5e9C) = 1.6e7 N/C`

If the test charge were -2.00 nC instead, the force on it would be as follows:

F = q*E = (-2*10⁻⁹ C) * 1.67*10⁷ N/C = -0.03 N

(The minus sign explains that the force is directed in opposite direction to the field, i.e., southwest).

b) The electric field is the same we just found above:

`E = (F)/(q) =(-0.0300N)/(-2e9C) = 1.6e7 N/C`

Answer 2

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

Step-by-step 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