Answer 6.2 a

_psets/4.md

@@ -48,6 +48,22 @@ Use Gauss’ Law to find the capacitance between two parallel plates of area A a
 difference V and with a spacing d. Neglect the fringing fields by assuming that this is a section of
 an infinite capacitor.
 
+The capacitance between two objects is the ratio of their charge to their potential difference.
+(This ratio does not depend on the charge because of the linearity of Maxwell's equations.) So to
+find the capacitance between two plates with a potential difference V, we need to know what charge
+is required to generate that potential difference.
+
+Consider a single infinite plane with a charge density $$\sigma$$. By symmetry the field must be
+perpendicular to the plane everywhere. For energy to be conserved this means that the field doesn't
+vary with distance! Consider a section of area $$A$$ on the plane. If we expand this symmetrically
+to a box that is bisected by the plane, Gauss' Law tells us that $$2 A f = \sigma A / \epsilon_0$$,
+so the field strength $$f$$ is $$\sigma / (2 \epsilon_0)$$.
+
+Thus between two planes with opposite charge densities $$\sigma$$ and $$-\sigma$$, the field is
+$$\sigma / \epsilon_0$$ everywhere. This means the potential difference between the planes is $$d
+\sigma / \epsilon_0$$. As such the capacitance is $$Q / V = A \sigma \epsilon_0 / (d \sigma) = A
+\epsilon_0 / d$$.
+
 ### (b)
 
 {:.question}