Edits to “Condition”, “Solution”, “Answer”

jzmicer edited
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revision #19196 parent #19195 ← older
en/6.5.6.md +39 −6
@@ -1,11 +1,44 @@
−### Statement
+### Condition
+$6.5.6$ Find the electric pressure on the inner surface of a spherical capacitor charged to a potential difference $V$. The radius of the outer plate of the capacitor is $R$, and the radius of the inner one is $r$.
−$6.5.6.$ [Insert the problem statement]
−
### Solution
+First, we prove the well‑known formula for the field pressure, which will be convenient to use in other problems:
+$$
+P=\frac{dF}{dS}=E\frac{dq}{dS}=\sigma E_{external}.
+$$
+For a conductor in vacuum, the field near its surface is related to the surface charge density by:
+$$
+\sigma=\varepsilon_0 E.
+$$
+This is easily obtained from Gauss's theorem, remembering that inside the conductor the field is zero.
−1
+Immediately near the surface, an area element behaves like an infinite plane. Such a plane creates its own field of $\sigma/(2\varepsilon_0)$ on each side. Then the external field is
+$$
+E_{external}=\frac{\sigma}{\varepsilon_0}-\frac{\sigma}{2\varepsilon_0}=\frac{E}{2}.
+$$
+From this we finally get:
+$$
+\boxed{p=\frac{\varepsilon_0 E^2}{2}}\tag{1}
+$$
−#### Answer
+In problem [6.4.5](https://savchenkosolutions.com/ru/6.4.5) the capacitance of a spherical capacitor is found:
+$$
+C=4\pi\varepsilon_0\frac{Rr}{R-r}\tag{2}
+$$
+Accordingly, the charge on each sphere is
+$$
+Q=VC=4\pi\varepsilon_0V\frac{Rr}{R-r}.
+$$
+The field near the inner sphere is
+$$
+E=\frac{Q}{4\pi\varepsilon_0r^2}=V\frac{R}{r(R-r)}\tag{3}
+$$
+$(3)\to(1)$
+$$
+P=\frac{\varepsilon_0R^2V^2}{2r^2(R-r)^2}.
+$$
−[Insert a concise answer or boxed result]
+#### Answer
+$$
+\boxed{P=\frac{\varepsilon_0R^2V^2}{2r^2(R-r)^2}}
+$$