Calculate ##\vec{E}## of a single charge at a certain distance from it

[mcastillo356]

Homework Statement

A single charge of $23\;\mu{C}$ is at the origin of a $x-y$ coordinate system. Calculate the electric field vector at the point $r=(1,25\;m,0,75\;m)$

Relevant Equations

$\vec{E}=K\dfrac{q}{r^2}$

Really don't know if this is wright. Here it goes. Thanks in advance:
$\vec{E}=K\cdot{q}\cdot{\sqrt{\dfrac{1}{r_1^4}+\dfrac{1}{r_2^4}}}\cdot{\left(\dfrac{r_1}{\sqrt{r_1^2+r_2^2}}\hat{i}+\dfrac{r_2}{\sqrt{r_1^2+r_2^2}}\hat{j}\right)}$
$8,98\cdot{10^{9}}\;Nm^2/C^2\cdot{23,6\cdot{10^{-6}}\;C}\cdot{\left(\sqrt{\frac{1}{2,44\;m^4}+\frac{1}{3,16\;m^4}}\right)}\left(\dfrac{1,25\;m}{1,46\;m}\hat{i}+\dfrac{0,75\;m}{1,46\;m}\hat{j}\right)$
$8,98\cdot{10^{9}}\;Nm^2/C^2\cdot{23,6\cdot{10^{-6}}\;C}\cdot{0,852\;m^{-2}}\left(0,856\hat{i}+0,514\hat{j}\right)$
$\therefore \vec{E}=1,55\cdot{10^5}\;N/C\hat{i}+0,928\cdot{10^5}\;N/C\hat{j}$
My doubt: how is it such a large number?
Greetings!

 

[haruspex]

I am unable to sort out your latex. You seem to be missing some curly braces and a \right).

 

[mcastillo356]

Homework Statement:: A single charge of $23\;\mu{C}$ is at the origin of a $x-y$ coordinate system. Calculate the electric field vector at the point $r=(1,25\;m,0,75\;m)$
Relevant Equations:: $\vec{E}=K\dfrac{q}{r^2}$

Really don't know if this is wright. Here it goes. Thanks in advance:
$\vec{E}=K\cdot{q}\cdot{\sqrt{\dfrac{1}{r_1^4}+\dfrac{1}{r_2^4}}}\cdot{\left(\dfrac{r_1}{\sqrt{r_1^2+r_2^2}}\hat{i}+\dfrac{r_2}{\sqrt{r_1^2+r_2^2}}\hat{j}\right)}$
$8,98\cdot{10^{9}}\;Nm^2/C^2\cdot{23,6\cdot{10^{-6}}\;C}\cdot{\left(\sqrt{\frac{1}{2,44\;m^4}+\frac{1}{3,16\;m^4}}\right)}\left(\dfrac{1,25\;m}{1,46\;m}\hat{i}+\dfrac{0,75\;m}{1,46\;m}\hat{j}\right)$
$8,98\cdot{10^{9}}\;Nm^2/C^2\cdot{23,6\cdot{10^{-6}}\;C}\cdot{0,852\;m^{-2}}\left(0,856\hat{i}+0,514\hat{j}\right)$
$\therefore \vec{E}=1,55\cdot{10^5}\;N/C\hat{i}+0,928\cdot{10^5}\;N/C\hat{j}$
My doubt: how is it such a large number?
Greetings!

I am quoting whith my phone. The pc did work

 

[PeroK]

I am quoting whith my phone. The pc did work

The numbers must be that order of magnitude as $\epsilon_0$ is very small. That said, don't you just put these numbers into a calculator and see what comes out? What do you gain by writing it all out like that?

I would calculate $r$, then use
$$\vec E = \frac{q}{4\pi \epsilon_0 r^3}\vec r$$
PS I get a different answer from you. You may have gone wrong inverting the value for $\epsilon_0$ by the look of your working.

 

[mcastillo356]

I thought I had to warn that I had solved my problems with LaTeX. I will not do it again. It's obvious I've not thought twice.

 

[PeroK]

I thought I had to warn that I had solved my problems with LaTeX. I will not do it again. It's obvious I've not thought twice.

What I meant was that once you have a formula and all the numbers, that's just one calculation on a calculator or spreadsheet. It's difficult to check all those intermediate numbers because I just put the numbers in a spreadsheet. Three calculations: one for $r$, one for $E_x$ and one for $E_y$:
$$\vec E = \frac{q}{4\pi \epsilon_0 r^3}(x, y) = \frac{q}{4\pi \epsilon_0 r^3}(x \vec i + y \vec j)$$

 

[mcastillo356]

Ahh...Yes, I've misunderstood. I must have missed some calculation