Physics, 01.08.2019 16:30 Aethis

Two particles, each carrying positive charge q, are on the y axis, one at y=+a and the other at y=−a. calculate electric field for any point on the x axis

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Answer from: ricksterv5000

The field for any point on the x-axis will determined as $E = \frac{kq}{4a^2+ x^2}$

The given parameters;

charge of each particle, = qthe position of one charge = + aposition of the second charge = -a

The electric field between these charge is determined by applying Coulomb's law as shown below;

$E = \frac{kq}{r^2}$

where;

r is the distance between the charges

The resultant distance at any point in the x-axis is calculated as follows;

$r^2 = (a- -a)^2 + (x)^2\\\\r^2 = (2a)^2 + (x)^2\\\\r^2 = 4a^2 + x^2$

The electric field is calculated as follows;

$E = \frac{kq}{r^2} = \frac{kq}{4a^2 + x^2}$

Thus, the field for any point on the x-axis will determined as $E = \frac{kq}{4a^2+ x^2}$

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Answer from: angelinaranee15

The electric field produced by a single particle of charge q is given by:
$E(r)=k_e \frac{q}{r^2}$
where k_e

is the Coulomb's constant, and r is the distance of the point we are measuring the field from the particle.

The problem asks to calculate the electric field for any point on the x-axis. Let's start with the charge located at y=+a. Let's take a generic point on the x-axis: this will have coordinates (x,0). Therefore, the distance of this point from the charge will be
$r= \sqrt{x^2+a^2}$
And so, if we substitute this into the formula of the electric field, we have
$E_1 = k_e \frac{q}{x^2+a^2}$

For the second charge, the electric field is the same. In fact, the second charge is also +q. It is located at y=-a, therefore the distance from the generic point x will be
$r= \sqrt{x^2+(-a)^2} = \sqrt{x^2+a^2}$
And so, the field generated by charge 2 is
$E_2=k_e \frac{q}{x^2+a^2}$

so, the total field is
$E_{tot}(x) = E_1 + E_2=2k_e \frac{q}{x^2+a^2}$

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

Your answer is 32 hope i

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Two particles, each carrying positive charge q, are on the y axis, one at y=+a and the other at y=−a...

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