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

• millahjallar
In summary, the conversation was about calculating electrostatic energy and the units involved. The dot product of D-field and E-field has units of V/m**2, and this is used in the formula W=0.5 integral_V(D dot E)dV to find the energy density in a field. The units of the integral will depend on the chosen differential element. It is important to use the correct units, such as the value for epsilon (ε0) in Farads/meter, which is not the same as the charge of an electron.
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?

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/

millahjallar
Hello @millahjallar ,
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] !

##\ ##

PeroK said:

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.. :)

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

millahjallar
millahjallar said:
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.

Orodruin said:
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.

millahjallar said:
I'm trying to calculate the electrostatic energy
Remind me what that dot product has to do with the electrostatic energy ...

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

BvU said:
Hello @millahjallar ,
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! :)

millahjallar said:
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
BvU said:
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.

BvU
millahjallar said:
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.
Are you using units where ##\epsilon_0 = 1##?

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##?

SammyS
millahjallar said:
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.

Last edited:
Mister T said:
##\frac{V^2}{m^4}## is not integrable. You need a differential element in there.
Poster is referring to the units

##\ ##

BvU said:
Poster is referring to the units

##\ ##
And the units depend on which differential element you choose.

PeroK said:
Are you using units where ##\epsilon_0 = 1##?
No, e = -1.60 :)

-1.60 C

millahjallar said:
Dot product: ##\vec{D} \cdot\vec{E}## in Si-units
millahjallar said:
-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 !

##\ ##

PeroK
millahjallar said:
No, e = -1.60 :)

SammyS
very interesting conversation, cleared my concept.

## 1. What is the definition of dot product in SI units?

The dot product of two vectors, ##\vec{D}## and ##\vec{E}##, is the product of their magnitudes and the cosine of the angle between them. In SI units, the dot product is measured in newton-meters (N·m).

## 2. How is the dot product calculated in SI units?

To calculate the dot product in SI units, first find the magnitudes of the two vectors. Then, multiply the magnitudes together and multiply the result by the cosine of the angle between the two vectors. The resulting unit will be N·m.

## 3. What is the physical significance of the dot product in SI units?

The dot product in SI units represents the work done by one vector on the other. It is also equal to the product of the two vectors' projections onto each other. In physics, the dot product is used to calculate the amount of force applied in a specific direction.

## 4. Can the dot product be negative in SI units?

Yes, the dot product can be negative in SI units. This occurs when the angle between the two vectors is greater than 90 degrees, resulting in a negative cosine value. A negative dot product indicates that the two vectors are pointing in opposite directions.

## 5. How is the dot product used in real-world applications?

The dot product has many real-world applications, such as calculating the torque applied to an object, finding the work done by a force on an object, and determining the angle between two vectors. It is also used in physics and engineering to solve problems involving forces, motion, and energy.

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