Location of charged particle given magnitude of position

In summary: The equation should 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...The equation would be ##\vec E= |\vec E|\hat E =\frac 1 {4π\varepsilon_0} \frac q {|\vec r|^2} \vec {E}##. Why does ##\hat r##/##\hat E## vanish from the equation? Wouldn't it
  • #1
Zack K
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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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  • #2
The magnitude of a vector ##\vec E## is given by ##|\vec E|=\sqrt{\vec E \cdot \vec E}.## Does this answer your question?
 
  • #3
kuruman said:
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.
 
  • #4
Zack K said:
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##.
 
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  • #5
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?
 
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  • #6
kuruman said:
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##
 
  • #7
Zack K said:
##|\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.
 
  • #8
Zack K said:
##\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}##
 
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  • #9
haruspex said:
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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  • #10
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
$$|\vec{E}|-\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2}=0\Rightarrow |\vec{E}|=\frac{1}{4\pi\epsilon_0}\frac{q}{|\vec{r}|^2}$$
 
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  • #11
Delta2 said:
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.
 
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FAQ: Location of charged particle given magnitude of position

1. What is the equation for calculating the location of a charged particle given its magnitude of position?

The equation for calculating the location of a charged particle is x = qE/m, where x is the position, q is the charge of the particle, E is the electric field, and m is the mass of the particle.

2. How do you determine the direction of the charged particle's location?

The direction of the charged particle's location can be determined by the direction of the electric field. The particle will move in the direction of the electric field lines.

3. Can the location of a charged particle change over time?

Yes, the location of a charged particle can change over time if the electric field or the particle's charge or mass changes. The particle will move in response to these changes.

4. How does the magnitude of the particle's position affect its movement?

The magnitude of the particle's position affects its movement by determining how far it will move in response to the electric field. A larger magnitude of position will result in a larger displacement.

5. Is the location of a charged particle affected by other particles in its vicinity?

Yes, the location of a charged particle can be affected by other particles in its vicinity if they also have a charge. The electric fields from these particles will interact and can change the direction and magnitude of the particle's movement.

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