Location of charged particle given magnitude of position

[Zack K]

Homework Statement

A charged particle has an electric field at $\langle -0.13, 0.14, 0 \rangle$ m is $\langle 6.48\times10^3, -8.64\times10^3, 0 \rangle$ N/C. The charged particle is -3nC. Where is the particle located?

Homework Equations

$\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|} \hat {r}$

The Attempt at a Solution

The solution was already given since this was a textbook example. Going through the process:
$|\vec E|=\sqrt{(6.4\times 10^3)^2 + (-8.64\times 10^3)^2} N/C = 1.08\times 10^4 N/C$
They then present the modified equation: $|\vec E|=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|}$
My question is why is that equation true? Why does substituting the vector $\vec E$ for the magnitude $|\vec E|$ get rid of the unit vector $\hat {r}$? The book never explained if they are related.

Edit: I didn't write down the entire solution since those steps make sense to me, it's only this part.

 

[kuruman]

The magnitude of a vector $\vec E$ is given by $|\vec E|=\sqrt{\vec E \cdot \vec E}.$ Does this answer your question?

 

[Zack K]

The magnitude of a vector $\vec E$ is given by $|\vec E|=\sqrt{\vec E \cdot \vec E}.$ Does this answer your question?

Isn't that the same thing as saying that $|\vec E|= \vec E?$ since $\vec E ⋅ \vec E = \vec E^2$ and $\sqrt {\vec E^2}=\vec E$. Or is this the case of a dot product, which we haven't covered yet.

 

[haruspex]

Isn't that the same thing as saying that $|\vec E|= \vec E?$ since $\vec E ⋅ \vec E = \vec E^2$ and $\sqrt {\vec E^2}=\vec E$. Or is this the case of a dot product, which we haven't covered yet.

When we write the square of a vector we necessarily mean its dot product with itself.
The √ function is defined to return a non-negative scalar. Even in scalars there is an ambiguity with taking a square root. √(x²)=|x|. If x²=y then x=±√y. Note that it is wrong to write √(x²)=x.
With vectors the corresponding ambiguity grows substantially. If |(x,y)|=c then all we can say is that x²+y²=c². The √ function is still defined to return a non-negative scalar, so $\sqrt {\vec x^2}=|\vec x|$, not $\vec x$.

 

[kuruman]

It's not the same. A vector has magnitude and direction. It is represented as $\vec E$. The magnitude of the vector is its size and has no direction. It is represented as $|\vec E|$ or more commonly as $E$. $\vec E \cdot \vec E=E^2$ is the square of the magnitude, it is a scalar and has no direction. It makes no sense to write $\sqrt{\vec E^2}$. What is the square root of a direction?

 

[Zack K]

It's not the same. A vector has magnitude and direction. It is represented as $\vec E$. The magnitude of the vector is its size and has no direction. It is represented as $|\vec E|$ or more commonly as $E$. $\vec E \cdot \vec E=E^2$ is the square of the magnitude, it is a scalar and has no direction. It makes no sense to write $\sqrt{\vec E^2}$. What is the square root of a direction?

Ah I see now that makes sense, but I'm still a little confused on how the substitution gets rid of $\hat r$. $|\vec E|=\frac {\vec E} {\hat E}$. So how exactly is it related to $\hat r$

 

[haruspex]

$|\vec E|=\frac {\vec E} {\hat E}$

No, no! $|\vec E| {\hat E}={\vec E}$, but thou shalt not divide by a vector - it's undefined.

 

[haruspex]

$\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|} \hat {r}$

I don't think you mean that. It should be either
$\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2} \hat {r}$
or
$\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^3} \vec {r}$
right?

This equation can be thought of as telling you two things.
First, that the direction of $\vec E$ is the same as that of $\vec r$, i.e. $\hat E=\hat r$.
Second, that the magnitude of $\vec E$ is $\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2}$.
Since $\vec E = |\vec E|\hat E$, these combine to produce $\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2} \hat {r}$

 

[Zack K]

I don't think you mean that. It should be either
$\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2} \hat {r}$
or
$\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^3} \vec {r}$
right?

This equation can be thought of as telling you two things.
First, that the direction of $\vec E$ is the same as that of $\vec r$, i.e. $\hat E=\hat r$.
Second, that the magnitude of $\vec E$ is $\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2}$.
Since $\vec E = |\vec E|\hat E$, these combine to produce $\vec E=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2} \hat {r}$

Sorry that was my bad, I forgot to square the bottom portion.
So if what you said is true. Shouldn't the equation be $\vec E= |\vec E|\hat E = \frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2} \hat {E}$? why does $\hat r$/$\hat E$ vanish from the equation? Wouldn't it be because you are dividing both sides by $\hat E$ to get $|\vec E|=\frac {\vec E} {\hat E}= \frac 1 {4π\varepsilon_0} \frac q {(|\vec r|^2)\hat E} \hat E$? I know you can't divide by vectors, but it seems to be the only way for me to cancel the unit vectors out and get $|\vec E|=\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2}$

 

[Delta2]

It does not vanish because of division by vector but because we move the right hand side to the left hand side and then take advantage of a property that if $\lambda \vec{v}=\vec{0}$ then either $\vec{v}=\vec{0}$ or $\lambda=0$
So what we do is:
$|\vec{E}|\hat{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2}\hat{E}\Rightarrow (|\vec{E}|-\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2})\hat{E}=\vec{0}$ and from the last equation and since we know that $\hat{E}\neq \vec{0}$ it follows that
$$|\vec{E}|-\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2}=0\Rightarrow |\vec{E}|=\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2}$$

 

[Zack K]

It does not vanish because of division by vector but because we move the right hand side to the left hand side and then take advantage of a property that if $\lambda \vec{v}=\vec{0}$ then either $\vec{v}=\vec{0}$ or $\lambda=0$
So what we do is:
$|\vec{E}|\hat{E}=\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2}\hat{E}\Rightarrow (|\vec{E}|-\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2})\hat{E}=\vec{0}$ and from the last equation and since we know that $\hat{E}\neq \vec{0}$ it follows that
$$|\vec{E}|-\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2}=0\Rightarrow |\vec{E}|=\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2}$$

I see now. Thank you.