Question

Finally, Place a positive (+) charge and a negative () charge on a horizontal line in the simulation. Make sure the arrows in-between the charges are horizontal. Then, place the sensor at a point in the simulation where the it is 250 cm from both the positive (+) and negative (-) charges. Let the sensor be taken as a positive (+) test charge, with a = +1 nC.

What is the magnitude of the Electric force at this point? 2.86 X N
Taking the positive x-axis to be horizontal and right, what is the angle as recorded by the sensor? 0.2
What are the x & y components of the Electric force? Fx = N Fy = N

Answer 1

Q = 2 10⁻² C

Step-by-step explanation:

This is an exercise where we must use that charges of the same sign repel and charges of a different sign attract, let's find the total force

F_total = F₁ + F₂

where the force F₁ is the force between the positive charge 1 and the positive test charge, so this force is to the right.

The force F₂ is the force between the negative charge 2 and the positive test charge, the direction is to the right

to find them force we use Coulomb's law

F₁ = k q₁ q / r²

F₂ = k q₂ q / r²

F_total = k q / r² (q₁ + q₂)

the charges q₁ and q₂ are equal

q₁ = q₂ = Q

F_total = k q Q / r²

In the problem we are given the distance r = 0.250 m, the Coulomb constant is k = 9 10⁹ N m² / C², the test charge q = 1 nC = 1 10⁻⁹ C, in order to calculate the net force we must know the charge Q, as is given, but the total force has a value of F_total = 2.86 N, we can find the load Q

Q = F_total r² / k q

we calculate

Q = 2.86 0.25² / (9 10⁹ 1 10⁻⁹)

Q = 1,956 10⁻² C

If the loads are on the horizontal line and they are uniformly distributed loads e, the sensor must measure an angle of ZERO, that is, the force must be horizontal.

From your answer you can see that the real angle is very small (0.2), so the initial assumption is correct.