Physics, 06.10.2020 01:01 61497

A charge of 0.5 μC is located at A (10,-10,10) (in cm) and a second charge of -0.3 μC is located at B (-10,15,5) (in cm). Find E (Electric Field) at P (15,20,50) (in cm).

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The electric field at E at point P is represented by $\vec E_{tot} = -3106.063\,\hat{i}+9694.512\,\hat{j}+5436.229\,\hat{k}\,\,\left[\frac{N}{C} \right]$ .

Explanation:

The definition of the electric field is represented by the following differential equation:

$\vec E = \frac{d\vec F}{dq_{o}}$

Where:

$\vec E$ - Electric field, measured in newtons per coulomb.

$\vec F$ - Electrostatic force, measured in newtons.

$q_{o}$ - Theoretical electric charge, measured in coulombs.

Electrostatic force is modelled by Coulomb's Law, which states that such force is directly proportional to electric current and inversely proportional to square distance. In this case, vectorial form of this law is:

$\vec F = \frac{\kappa\cdot q\cdot q_{o}}{r^{3}}\cdot \vec r$

Where:

$\kappa$ - Electrostatic constant, measured in newton-square meters per square coulomb.

- Real electric charge, measured in coulombs.

- Magnitude of the vector position, measured in meters.

$\vec r$ - Vector position, measured in meters.

We get the electric field equation by deriving the equation above:

$\vec E = \frac{\kappa\cdot q}{r^{3}}\cdot \vec r$

The total electric field is obtained by means of this concept and the Principle of Superposition as well:

$\vec{E}_{tot} = \vec {E}_{A}+\vec E_{B}$

$\vec {E}_{tot} = \frac{\kappa\cdot q_{A}}{r^{3}_{P/A}}\cdot \vec {r}_{P/A} + \frac{\kappa\cdot q_{B}}{r^{3}_{P/B}}\cdot \vec {r}_{P/B}$

At first we determine each vector position and their magnitudes:

$\vec r_{P/A} = \vec r_{P}-\vec r_{A}$

$\vec r_{P/A} = 0.15\,\hat{i}+0.20\,\hat{j}+0.50\,\hat{k}-0.10\,\hat{i}+0.10\,\hat{j}-0.10\,\hat{k}\,\,[m]$

$\vec r_{P/A} = 0.05\,\hat{i}+0.30\,\hat{j}+0.40\,\hat{k}\,\,[m]$

$r_{P/A} = \sqrt{(0.05\,m)^{2}+(0.30\,m)^{2}+(0.40\,m)^{2}}$

$r_{P/A} \approx 0.502\,m$

$\vec r_{P/B} = \vec r_{P}-\vec r_{B}$

$\vec r_{P/B} = 0.15\,\hat{i}+0.20\,\hat{j}+0.50\,\hat{k}+0.10\,\hat{i}-0.15\,\hat{j}-0.05\,\hat{k}\,\,[m]$

$\vec r_{P/B} = 0.25\,\hat{i}+0.05\,\hat{j}+0.45\,\hat{k}\,\,[m]$

$r_{P/B} = \sqrt{(0.25\,m)^{2}+(0.05\,m)^{2}+(0.45\,m)^{2}}$

$r_{P/B} \approx 0.517\,m$

If we know that $\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{A} = 0.5\times 10^{-6}\,C$ , $q_{B} = -0.3\times 10^{-6}\,C$ , $\vec r_{P/A} = 0.05\,\hat{i}+0.30\,\hat{j}+0.40\,\hat{k}\,\,[m]$ , $r_{P/A} \approx 0.502\,m$ , $\vec r_{P/B} = 0.25\,\hat{i}+0.05\,\hat{j}+0.45\,\hat{k}\,\,[m]$ and $r_{P/B} \approx 0.517\,m$ , then we get the electric field at point P.

$\vec{E}_{tot} = \left[\frac{\left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (0.5\times 10^{-6}\,C)}{(0.502\,m)^{3}} \right]\cdot (0.05\,\hat{i}+0.30\,\hat{j}+0.40\,\hat{k})\,\,[m] + \left[\frac{\left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (-0.3\times 10^{-6}\,C)}{(0.517\,m)^{3}} \right]\cdot (0.25\,\hat{i}+0.05\,\hat{j}+0.45\,\hat{k})\,\,[m]$ $\vec E_{tot} = 1778.572\,\hat{i}+10671.439\,\hat{j}+14228.573\,\hat{k} - 4884.635\,\hat{i}-976.927\,\hat{j}-8792.344\,\hat{k}\,\,\left[\frac{N}{C} \right]$ $\vec E_{tot} = -3106.063\,\hat{i}+9694.512\,\hat{j}+5436.229\,\hat{k}\,\,\left[\frac{N}{C} \right]$

The electric field at E at point P is represented by $\vec E_{tot} = -3106.063\,\hat{i}+9694.512\,\hat{j}+5436.229\,\hat{k}\,\,\left[\frac{N}{C} \right]$ .

Answer from: Quest

Mfjh f fb gcbb etyb yryh htbgh g yrjjj t n j tj

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.

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A charge of 0.5 μC is located at A (10,-10,10) (in cm) and a second charge of -0.3 μC is located at...

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