Edit to “Solution”

Luisito edited
revision #18528 parent #18527 ← older newer →
@@ -24,7 +24,7 @@Solution
developing,\
$V(r) = \frac{Q}{4\pi\varepsilon_0 R} - \frac{Q}{8\pi\varepsilon_0 R^3}(r^2-R^2)$\
as $r$ tends to zero,\
−$V(r) = \frac{3Q}{8\pi\varepsilon_0 R^3}$ (3)\
+$V(0) = \frac{3Q}{8\pi\varepsilon_0 R^3}$ (3)\
Case 2) Conducting sphere\
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),\
$V(r) = -\int_{\infty}^{R} \frac{Q}{4\pi\varepsilon_0 r^2}dr$\
unchanged lines 6