Physics, 06.04.2020 03:00 kingnatsu8638

A positive charge Q1 = 30.5 nC is located at the origin. A negative charge Q2 = -3.5 nC is located on the positive x-axis p = 3 cm from the origin. calculate the location, x on the x-axis in centimeters where the electric field is equal to zero

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Answer from: kelly1027

The points where the electric field due to the two charges are 0 include

x = 4.54 cm and x = 2.24 cm.

Explanation:

For a positive charge, the electric field is directed outward from the charge,

For a negative charge, it is directed towards it.

Since the positive charge is at the origin, the, and the negative charge is at p = 3cm on the x-axis, let the point where the electric field is 0 be x.

For ease of calculation, we will be looking for x, along the positive x-axis in between the two charges.

Electric field at a point due to a particular charge is given as

E = (kq/r²)

k = Coulomb's constant = (9.0 × 10⁹) Nm²/C²

r = distance of the point from the charge.

q = the charge

q₁ = 30.5 nC = (30.5 × 10⁻⁹) C

q₂ = -3.5 nC = (3.5 × 10⁻⁹) C

For the positive charge

E₁ = (kq₁)/x²

= (9.0 × 10⁹ × 30.5 × 10⁻⁹)/x²

E₁ = (274.5/x²) (eqn 1)

For the negative charge

E₂ = [kq₂/(x-0.03)²]

E₂ = 31.5/(x-0.03)²

E₁ + E₂ = 0

(274.5/x²) - [31.5/(x-0.03)²] = 0 (with the assumption that the point is to the right of both charges)

(274.5/x²) =-[31.5/(x-0.03)²]

274.5 (x-0.03)² = 31.5x²

274.5 (x² - 0.06x + 0.0009) = 31.5x²

274.5x² - 16.47x + 0.24705 = 31.5x²

243x² - 16.47x + 0.24705 = 0

Solving this equation

x = 0.0454 m or x = 0.0224 m

So, the points where the electric field due to the two charges are 0 include

x = 4.54 cm and x = 2.24 cm.

Hope this Helps!!!

Answer from: Quest

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

$p(t)=0.3t^2+6t+80$

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

so, our quadratic equation becomes,

$p(0)=0+0+80=80$

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

$p(t)=2\times 80=160$

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

$160=0.3t^2+6t+80\\\\160-80=0.3t^2+6t\\\\80=0.3t^2+6t\\\\0.3t^2+6t-80=0\\\\\text{using quadratic formula ,we get }\\\\t_{1,\: 2}=\frac{-60\pm \sqrt{60^2-4\cdot \: 3\cdot \: 800}}{2\cdot \: 3}\\\\t=-29.14\ and\ t=9.149$

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.

Answer from: Quest

e/c, where e is its total energy

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A positive charge Q1 = 30.5 nC is located at the origin. A negative charge Q2 = -3.5 nC is located o...

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