- #1
neutrino
- 2,091
- 2
Griffiths : Electrostatic Energy
I'm having a little difficulty in understanding how one arrives at the following expression for electrostatic energy of a continuous charge distribution.
W = \frac{\epsilon_o}{2}\int (\vec{E})^2d\tau
This result is obtained when the volume of integration is increased without limit in
W = \frac{\epsilon_o}{2} ( \int (\vec{E})^2d\tau + \oint V \vec{E} \cdot d\vec{a})
For large ditances from the charge V goes like \frac{1}{r} while \vec{E} goes like \frac{1}{r^2}, and the area increases as r^2. Therefore the surface integral is roughly \frac{1}{r}. Griffiths says that the volume integral must increase owing to the positive integrand. But can't we apply similar logic as above and say that volume integral also rougly goes like \frac{1}{r} as we increase of the volume of integration?
I'm having a little difficulty in understanding how one arrives at the following expression for electrostatic energy of a continuous charge distribution.
W = \frac{\epsilon_o}{2}\int (\vec{E})^2d\tau
This result is obtained when the volume of integration is increased without limit in
W = \frac{\epsilon_o}{2} ( \int (\vec{E})^2d\tau + \oint V \vec{E} \cdot d\vec{a})
For large ditances from the charge V goes like \frac{1}{r} while \vec{E} goes like \frac{1}{r^2}, and the area increases as r^2. Therefore the surface integral is roughly \frac{1}{r}. Griffiths says that the volume integral must increase owing to the positive integrand. But can't we apply similar logic as above and say that volume integral also rougly goes like \frac{1}{r} as we increase of the volume of integration?
Last edited:
)
