Recent content by millahjallar

  1. M

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

    -1.60 C
  2. M

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

    No, e = -1.60 :)
  3. M

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

    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! :)
  4. M

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

    Okay, thank you! :)
  5. M

    Dot product: ##\vec{D} \cdot\vec{E}## in SI 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.
  6. M

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

    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.
  7. M

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

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

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

    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?