Physics, 21.03.2020 22:24 ssiy

Una carga de -3 µC y otra de 2 µC en el vacío están separadas 4 cm.
A. Calcula la magnitud de la fuerza electrostática que ejerce una de las cargas sobre la otra.
B. Dibuja la fuerza electrostática que actúa sobre cada carga.
C. Calcula la magnitud de la fuerza electrostática si las cargas estuvieran en vidrio.

Answers: 3

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Answers

Answer from: justabeachbum

A) 33.8 N

B) Attractive force (see graph)

C) 158.9 N

Explanation:

The magnitude of the electrostatic force between two electric charges is given by

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the magnitudes of the two charges

r is the separation between the charges

Here in this problem, we have:

$q_1 =3 \mu C=3\cdot 10^{-6}C$ is the magnitude of the 1st charge

$q_2=2 \mu C=2\cdot 10^{-6}C$ is the magnitude of the 2nd charge

r = 4 cm = 0.04 m is the separation

Substituting, we find the magnitude of the force:

$F=\frac{(9\cdot 10^9)(3\cdot 10^{-6})(2\cdot 10^{-6})}{(0.04)^2}=33.8 N$

Find the graph showing the direction of the force in attachment.

The direction of the electrostatic force between two charges is given as follows:

- If the two charges have same sign (++ or --), the force is repulsive

- If the two charges have opposite sign (+-), then the force is attractive

In this problem:

- Charge 1 is negative

- Charge 2 is positive

The two charges have opposite signs: therefore, according to the rules above, the electrostatic force between them is attractive.

When the charges are placed in a medium, the magnitude of the electrostatic force between the two charges increases due to the polarization effect on the medium, and the new force is given by

$F'=\epsilon_r F$

where:

F is the original electrostatic force in a vacuum

$\epsilon_r$ is the relative permittivity of the medium

Here in this problem, we have:

F = 33.8 N is the original electrostatic force

$\epsilon_r = 4.7$ is the relative permittivity of the medium (glass)

Therefore, the new electrostatic force between the charges is

F' = (33.8)(4.7) = 158.9 N

Una carga de -3 µC y otra de 2 µC en el vacío están separadas 4 cm. A. Calcula la magnitud de la fue

Answer from: Quest

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Answer from: Quest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .