Find the potential inside and outside a spherical shell of radius R (Fig. 2.31) that carries a uniform surface charge. Set the reference point at infinity.
Question 2.7: Find the potential inside and outside a spherical shell of r...
From Gauss’s law, the field outside is
E=\frac{1}{4\pi \epsilon _{0}}\frac{q}{r^{2}}\acute{r}.where q is the total charge on the sphere. The field inside is zero. For points outside the sphere (r > R),
V(r)=-\int_{\omicron }^{r}{E.dI}=\frac{-1}{4\pi \epsilon _{0}}\int_{\infty }^{r}{\frac{q}{\acute{r}^{2}} }d\acute{r} = \frac{1}{4\pi \epsilon _{0}}\frac{q}{\acute{r}}\mid ^{r}_{\infty } =\frac{1}{4\pi \epsilon _{0}}\frac{q}{r}.To find the potential inside the sphere (r < R), we must break the integral into two pieces, using in each region the field that prevails there:
V(r)=\frac{-1}{4\pi \epsilon _{0}}\int_{\infty }^{R}{\frac{q}{\acute{r}^{2}} }d\acute{r}\int_{R}^{r}{(0)d\acute{r}} = \frac{1}{4\pi \epsilon _{0}}\frac{q}{\acute{r}}\mid ^{R}_{\infty } +0=\frac{1}{4\pi \epsilon _{0}}\frac{q}{R}.Notice that the potential is not zero inside the shell, even though the field is.V is a constant in this region, to be sure, so that ∇V = 0—that’s what matters.In problems of this type, you must always work your way in from the reference point; that’s where the potential is “nailed down.” It is tempting to suppose that you could figure out the potential inside the sphere on the basis of the field there alone, but this is false: The potential inside the sphere is sensitive to what’s going on outside the sphere as well. If I placed a second uniformly charged shell out at radius \acute{R}\gt R, the potential inside R would change, even though the field would still be zero. Gauss’s law guarantees that charge exterior to a given point that is, at larger r ) produces no net field at that point, provided it is spherically or cylindrically symmetric, but there is no such rule for potential, when infinity is used as the reference point.
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