Dot product: ##\vec{D} \cdot\vec{E}## in SI units

[millahjallar]

Homework Statement

Electromagnetism - physics
Dot product: $\vec{D} \cdot\vec{E}$ in Si-units

Relevant Equations

D dot E

I'm trying to calculate the electrostatic energy, and I'm wondering what happens when I dot the D-field and E-field, with Si-units V/m**2. This is my equation:

D dot E = (-4x(epsilon) V/m**2)(-4x V/m**2) + (-12y(epsilon) V/m**2)(-12y V/m**2)

Are the final Si-unit still V/m**2 or V**2/m**4?

 

[PeroK]

The dot product is the vector equivalent of multiplication, so the units multiply as well. E.g. $$dW = \vec F \cdot d\vec {r}$$ has units of energy.

For future reference you might like to learn some Latex to render your equations:

https://www.physicsforums.com/help/latexhelp/

 

[BvU]

Hello @millahjallar ,

$\qquad$!
​

The units of $\vec D\cdot\vec E$ are the product of the units of $D$ and $E$.
The units of $\varepsilon$ are not [1] !

$\$

 

[millahjallar]

The dot product is the vector equivalent of multiplication, so the units multiply as well. E.g. $$dW = \vec F \cdot d\vec {r}$$ has units of energy.

For future reference you might like to learn some Latex to render your equations:

https://www.physicsforums.com/help/latexhelp/

Okay, great - thanks!
I should definitely learn how to write in latex.. :)

 

[Orodruin]

Note: The dot product itself is a mathematical construct and not related to any particular set of units.

 

[millahjallar]

Okay, great - thanks!
I should definitely learn how to write in latex.. :)

If I want to take the triple-integral of V^2/m^4, does the unit change?
Again - sorry for not currently knowing how to write in latex.

 

[millahjallar]

Note: The dot product itself is a mathematical construct and not related to any particular set of units.

Okay, thank you! But if I want to take the triple-integral of V^2/m^4, does the unit change?
Sorry for not currently knowing how to write in latex.

 

[BvU]

I'm trying to calculate the electrostatic energy

Remind me what that dot product has to do with the electrostatic energy ...

if I want to take the triple-integral

A triple integral of something is usually over the volume. The integration adds a factor of $m^3$ to whatever is integrated

$\$

 

[millahjallar]

Okay, great - thanks!
I should definitely learn how to write in latex.. :

Hello @millahjallar ,

$\qquad$!
​

The units of $\vec D\cdot\vec E$ are the product of the units of $D$ and $E$.
The units of $\varepsilon$ are not [1] !

$\$

Okay, thank you! :)

 

[PeroK]

If I want to take the triple-integral of V^2/m^4, does the unit change?
Again - sorry for not currently knowing how to write in latex.

For an integral, the integrand is essentially multiplied by the variable over which integration takes place: displacement is the area under a velocity against time graph: metres/second times seconds give metres. Mass is the integral of density (mass per unit volume) integrated over a volume. Etc.

 

[millahjallar]

Remind me what that dot product has to do with the electrostatic energy ...A triple integral of something is usually over the volume. The integration adds a factor of $m^3$ to whatever is integrated

$\$

I want to find the electrostatic energy in a field, and in order to do so, I'm using the formula W=0.5 integral_V(D dot E)dV.
Thank you so much for your reply! :)

 

[PeroK]

I want to find the electrostatic energy in a field, and in order to do so, I'm using the formula W=0.5 integral_V(D dot E)dV.
Thank you so much for your reply! :)

Are you using units where $\epsilon_0 = 1$?

 

[kuruman]

The energy density $u$ (= energy per unit volume $V$) is given by $u=\frac{1}{2}\vec D\cdot\vec E$. What would be the dimensions of the integral $\int \frac{1}{2}\vec D\cdot\vec E~dV$?

 

[Mister T]

Okay, thank you! But if I want to take the triple-integral of V^2/m^4, does the unit change?

$\frac{V^2}{m^4}$ is not integrable. You need a differential element in there.

 

[BvU]

$\frac{V^2}{m^4}$ is not integrable. You need a differential element in there.

Poster is referring to the units

$\$

 

[Mister T]

Poster is referring to the units

$\$

And the units depend on which differential element you choose.

 

[millahjallar]

Are you using units where $\epsilon_0 = 1$?

No, e = -1.60 :)

 

[millahjallar]

-1.60 C

 

[BvU]

Dot product: $\vec{D} \cdot\vec{E}$ in Si-units

-1.60 C

This makes no sense at all. There is a world of difference between $e = 1.60217662 \times 10^{-19}$ Coulomb and $\varepsilon_0 = 8.8541878128 10^{-12}$ Farad/meter ( or kg^-1 m^-3 s⁴ A² ).

$e$ doesn't even occur in the exercise !

$\$

 

[kuruman]

No, e = -1.60 :)

Please read and answer the question in post #13.

 

[salmanshu322]

very interesting conversation, cleared my concept.