# 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.

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

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.

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##.

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?

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##

##|\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.

##\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}##

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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.

Delta2