Electric charge inside a cubical volume

AI Thread Summary
The discussion revolves around calculating the total electric charge within a cubical volume using Gauss's Law. The electric field intensity is given as E = 100x (V/m), and the user is confused about the units in their calculations. They correctly apply Gauss's Law and the Divergence Theorem but misinterpret the units of the divergence of the electric field. Clarification is provided that the divergence is in V/m², which, when multiplied by the volume, results in the correct unit of charge in coulombs. The user is reassured that their calculations are fundamentally correct, and the confusion stems from unit interpretation.
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Would someone be so kind to check if I am doing this properly. I'm confused on the units, as it doesn't seem to be coming out properly.

Q) Assuming that the electric field intensity is \vec E = \hat x 100 x \,\,(V/m), find the total electric charge contained inside a cubical volume 100 \,\, (mm) on a side centered symmetrically at the orgin.

My Work)
Recall:

\oint \vec E \cdot d\vec s = \frac{Q}{\epsilon_0} (Gauss's Law)
\int_V \nabla \cdot \vec A \, dv = \oint_S \vec A \cdot d\vec s (Divergence Thm)

Thus,
\oint_S \vec E \cdot d\vec s = \int_V \nabla \cdot \vec E \, dv = \frac{Q}{\epsilon_0}

\nabla \cdot \vec E = 100 \,\, (V/m)
100 (V/m) \int_V \, dv = 100\, (V/m)(100\times 10^{-3})^3(m^3) = \frac{1}{10} \,\, (V/m^2)

\frac{1}{10} \,\, (V/m^2) = \frac{Q}{\epsilon_0}

Thus,
Q = \frac{\epsilon_0}{10} \,\, (v/m^2) = 8.854\times 10^{-12} \frac{coul}{m^3}

I thought the units for Q should be in coul? Why am I getting coul per unit volume? Am I not doing this right?
 
Last edited:
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Hi, it looks like a units problem, the numbers look fine.

\nabla \cdot \vec E is measured in V/m2, not V/m. Then you multiply by m3, and get Vm. The units of {\epsilon_0} are C/Vm, so Vm cancels and you get C.
 
Mighty appreciated my man :smile:

thanks
 
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