Physics, 04.04.2021 19:30 Felixthecat8241

Three charged particles of charges 3 μC, -2 μC, and 4 μC are placed on the X-Y plane at (1 cm, 0), (2.5 cm, 0), and (1 cm, 2 cm) respectively. Determine the magnitude and direction of the force acting on a -2 μC charge.

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Answer from: aliami0306oyaj0n

The total force is $F_{tot}=4.29\: N$

The direction is $\omega=32.43^{\circ}$

Explanation:

First, we need to find the angle with respect to the horizontal, of the force between q2 (-2 μC) and q3 (3 μC).

Let's use the tangent function.

$tan(\alpha)=\frac{2}{1.5}$

$\alpha=53.13^{\circ}$

Now, let's find the magnitude of the force F(12).

$|F_{12}|=k\frac{q_{1}q_{2}}{d_{12}}$

Where:

k is the Coulomb constant (9*10⁹ NC²/m²)q1 is 3 μCq2 -2 μCd(12) is the distance between q1 and q2 ( 1.5 cm = 0.015 m)

$|F_{12}|=9*10^{9}\frac{3*10^{-6}*2*10^{-6}}{0.015}$

$|F_{12}|=3.6\: N$

The magnitude of the force F(23) will be:

$|F_{23}|=k\frac{q_{2}q_{3}}{d_{23}}$

The distance between these charges is:

$d_{23}=\sqrt{1.5^{2}+2^{2}}$

$d_{23}=2.5\: m$

$|F_{23}|=9*10^{9}\frac{2*10^{-6}*4*10^{-6}}{0.025}$

$|F_{23}|=2.88\: N$

So, we have the force F(12) in the second quadrant and F(23) in the second quadrant too but with 53.13 ° with respect to the horizontal.

We just need to add these two forces (vectors) and get the total force acting on q2.

Total force in x-direction:

$F_{tot-x}=-F_{12}-F_{23}cos(53.13)$

$F_{tot-x}=-3.6-2.88cos(53.13)$

$F_{tot-x}=-5.33\: N$

Total force in y-direction:

$F_{tot-y}=F_{23}sin(53.13)$

$F_{tot-y}=2.88sin(53.13)$

$F_{tot-y}=2.3\: N$

Therefore, the magnitude of the total force will be:

$|F_{tot}|=\sqrt{(-3.62)^{2}+(2.3)^{2}}$

$|F_{tot}|=4.29\: N$

and the direction is:

$tan(\omega)=\frac{2.30}{3.62}$

$\omega=32.43^{\circ}$

I hope it helps you!

Answer from: Quest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

Answer from: Quest

well clustered together.

points close on a scatter diagram

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