- #1
ozone
- 121
- 0
We were given an electric field defined by Kr^3, and asked to calculate what the total flux would be given a sphere of a radius R. I had already calculated the divergence of E to be equal to 5kr^2. So the first integral is calculating what the divergence over the area of the sphere is equal to, and our second integral is calculating what the surface integral of E * da is. By divergence theorem they should be equal but for me they are off by a factor of two.
\int_{\vartheta=0}^{2\pi} \int_{<br /> \varphi=0}^{\pi/2}\int_{\rho=0}^{R} 5Kr^2 (r^2 sin(\phi) d\rho d\phi d\theta
which will then yield 2K\pi R^5
Here is the surface integral.
\oint\oint E * da
where da will yield (4\pi R^2)
multiplying this times KR^3 gives us 4K\pi R^5
Any help on what to do here will be greatly appreciated.
THank you
\int_{\vartheta=0}^{2\pi} \int_{<br /> \varphi=0}^{\pi/2}\int_{\rho=0}^{R} 5Kr^2 (r^2 sin(\phi) d\rho d\phi d\theta
which will then yield 2K\pi R^5
Here is the surface integral.
\oint\oint E * da
where da will yield (4\pi R^2)
multiplying this times KR^3 gives us 4K\pi R^5
Any help on what to do here will be greatly appreciated.
THank you