Question

Given a polarized sphere with
PT) = Art
and electric field

Determine potential and electrostatic energy.

Can you do the potential using the line integral of the electric field. i have some doubts how to do the math.

Answer 1

we have electric field

and we do not need a polarisation because we have already a electric field

moglo E() = Ar --fir < R Eo or > R

so for potential

$V=\int E.dr$

where V= electric potential and E = electric field

now

$V=\int_{\infty }^{R}E_{1}dr + \int_{R}^{r}E_{2}dr$

$V=-\left ( \int_{\infty }^{R}0.dr +\int_{R}^{r}\frac{-Ar}{\epsilon _{0}}.dr \right )$

$V= 0 + \frac{A}{\epsilon _{0}}\int_{R}^{r}rdr$

$V= \frac{A}{2\epsilon _{0}}(r^{2}-R^{2})$

FOR electrostatic energy

$W= \frac{\epsilon _{0}}{2}\int \int \int E^{2}d\tau$

for spherical surface $d\tau = r^{2}\sin \theta drd\theta d\phi$

so

$W= \frac{\epsilon _{0}}{2}\int_{0}^{r}\int_{0}^{2\pi }\int_{0}^{\pi }E^{2}d\tau$

$W= \frac{\epsilon _{0}}{2}\int_{0}^{r}\int_{0}^{2\pi }\int_{0}^{\pi }E^{2}r^{2}sin\theta drd\theta d\phi$

SO NOW

$W=\frac{\epsilon _{0}}{2}\int_{0}^{R}\int_{0}^{2\pi }\int_{0}^{\pi }E_{1}^{2}r^{2}sin\theta drd\theta d\phi +\frac{\epsilon _{0}}{2}\int_{R}^{\infty }\int_{0}^{2\pi }\int_{0}^{\pi }E_{2}^{2}r^{2}sin\theta drd\theta d\phi$

$W=\frac{\epsilon _{0}}{2}\int_{0}^{R}\int_{0}^{2\pi }\int_{0}^{\pi }\left ( \frac{-Ar}{\epsilon _{0}}) \right )^{2}r^{2}sin\theta drd\theta d\phi +\frac{\epsilon _{0}}{2}\int_{R}^{\infty }\int_{0}^{2\pi }\int_{0}^{\pi }0.r^{2}sin\theta drd\theta d\phi$

$W = \frac{\epsilon _{0}}{2} \left ( \frac{A^{2}}{\epsilon _{0}^{2}} \right )\int_{0}^{R}r^{2}r^{2}dr\int_{0}^{\pi }sin\theta d\theta \int_{0}^{2\pi }d\phi$

$W = \frac{\epsilon _{0}}{2} \left ( \frac{A^{2}}{\epsilon _{0}^{2}} \right ) \left ( \frac{R^{5}}{5} \right )(2)(2\pi)$

$W =\left ( \frac{2\pi R^5A^{2}}{5\epsilon _{0}} \right )$