Правка раздела «Solution»

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@@ -29,8 +29,8 @@Solution
In this case, charge stays on sphere surface, so the electric field inside of it is null ($E(r) = 0$), so potential is constant inside the sphere and coincides with the value of it on the surface. According to (1) and (2),\
$V(r) = -\int_{\infty}^{R} \frac{Q}{4\pi\varepsilon_0 r^2}dr$\
$V(0) = V(r) = \frac{Q}{4\pi\varepsilon_0 R}$ (4)\
−Finally, if sphere is non-conductive $V(0)$ depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it **doesn't depend** on it. If the charge is non-uniformly distributed, the potential on the surface **does change** with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge, Q, as if it were a point charge at the center.
+Finally, if sphere is non-conductive $V(0)$ depends on the charge distribution, as we saw in (I), but in the second case (maybe the problem refers specifically to a conducting sphere), it **doesn't depend** on it. If the charge is non-uniformly distributed, the potential on the surface **does change** with the local distribution. For example, charge accumulations in certain areas generate angular variations in the potential. However, outside the sphere, at a great distance, the potential depends only on the total charge $Q$, as if it were a point charge at the center.
#### Answer
$V(0) = \frac{Q}{4\pi\varepsilon_0 R}$, NO, YES.