- #1
amcavoy
- 665
- 0
Let's say that the electric field near the surface of a sphere or radius R is E. I need to show that the energy density D is
[tex]D=\frac{\epsilon_{0}e^{2}}{2}.[/tex]
I started by noting that [itex]U=\frac{1}{2}Q\Delta V=\frac{kQ^{2}}{2R}[/itex]. The volume of the sphere is [itex]\frac{4}{3}\pi R^{3}[/itex]. This is how I found the energy density:
[tex]D=\frac{U}{\textrm{Vol}}=\frac{\left(\frac{kQ^{2}}{2R}\right)}{\left(\frac{4}{3}\pi R^{3}\right)}=\frac{3\epsilon_{0}E^{2}}{2}.[/tex]
Where is that extra 3 coming from? I can't really see where I went wrong (unless I used volume instead of surface area, in which case my answer would still be different).
Thanks.
[tex]D=\frac{\epsilon_{0}e^{2}}{2}.[/tex]
I started by noting that [itex]U=\frac{1}{2}Q\Delta V=\frac{kQ^{2}}{2R}[/itex]. The volume of the sphere is [itex]\frac{4}{3}\pi R^{3}[/itex]. This is how I found the energy density:
[tex]D=\frac{U}{\textrm{Vol}}=\frac{\left(\frac{kQ^{2}}{2R}\right)}{\left(\frac{4}{3}\pi R^{3}\right)}=\frac{3\epsilon_{0}E^{2}}{2}.[/tex]
Where is that extra 3 coming from? I can't really see where I went wrong (unless I used volume instead of surface area, in which case my answer would still be different).
Thanks.