Charge 1 is -30 μC, charge 2 is 5 μC, the distance between the charges is 8 cm, and they are located in a vacuum. If the third charge of 3 μC would be placed into position B, i. e., exactly between the two other charges, what would be the magnitude of the net force it experiences?

F₃ = 590.625 N

Explanation:

We have a situation where, we have 3 Charges, as follows:

Charge 1 = q₁ = - 30 μC

Charge 2 = q₂ = 5 μC

Charge 3 = q₃ = 3 μC

Distance between Charge 3 and Charge 1 = r₃₁ = 4 cm = 0.04 m

Distance between Charge 3 and Charge 2 = r₃₂ = 4 cm = 0.04 m

So, in order to find net force on the charge 3:

F₃ = F₁₃ + F₂₃

where,

F₃ = Net force on charge 3

F₁₃ = Force on Charge 3 due to Charge 1

F₂₃ = Force on Charge 3 due to Charge 2

We are adding the magnitudes of forces, because their direction will be same. Since, charge 1 is negative and charge 3 is positive, so direction of force will be towards charge 1. And charge 2 is positive thus direction of its force will also be towards charge 1.

Now, using Coulomb Law:

F₃ = kq₁q₃/r₁₃² + kq₃q₂/r₁₂²

F₃ = kq₃ (q₁/r₁₃² + q₂/r₁₂²)

F₃ = (9 x 10⁹ N.m²/C²)(3 x 10⁻⁶ C)[(30 x 10⁻⁶ C)/(0.04 m)² + (5 x 10⁻⁶ C)/(0.04 m)²

F₃ = 590.625 N

answer: at 24 february, 2014 the population gets twice of its initial population.

explanation:

since we have given that

the population for alpha city, t years after january 1, 2004 is given as

first we find out the initial population, i.e. t=0,

so, our quadratic equation becomes,

according to question, we have said that if the population has twice its initial population, then it becomes

so, time taken to reach this above mentioned population is given by

now, we know that time can't be negative so, we take t=9.149 years.

hence, at 24 february, 2014 the population gets twice of its initial population.