Simple Nature - Light and Matter

self-check DWe can think of the quantity dU e / dv as the energy density due to theelectric field, i.e., the number of joules per cubic meter needed in orderto create that field. (a) How does this quantity depend on the componentsof the field vector, E x , E y , and E z ? (b) Suppose we have a fieldwith E x ≠ 0, E y =0, and E z =0. What would happen to the energy densityif we reversed the sign of E x ? ⊲ Answer, p. 925A numerical example example 14⊲ A capacitor has plates whose areas are 10 −4 m 2 , separated bya gap of 10 −5 m. A 1.5-volt battery is connected across it. Howmuch energy is sucked out of the battery and stored in the electricfield between the plates? (A real capacitor typically has aninsulating material between the plates whose molecules interactelectrically with the charge in the plates. For this example, we’llassume that there is just a vacuum in between the plates. Theplates are also typically rolled up rather than flat.)⊲ To connect this with our previous calculations, we need to findthe charge density on the plates in terms of the voltage we weregiven. Our previous examples were based on the assumption thatthe gap between the plates was small compared to the size of theplates. Is this valid here? Well, if the plates were square, thenthe area of 10 −4 m 2 would imply that their sides were 10 −2 m inlength. This is indeed very large compared to the gap of 10 −5m, so this assumption appears to be valid (unless, perhaps, theplates have some very strange, long and skinny shape).Based on this assumption, the field is relatively uniform in thewhole volume between the plates, so we can use a single symbol,E, to represent its magnitude, and the relation E = dV / dx isequivalent to E = ∆V /∆x = (1.5 V)/(gap) = 1.5 × 10 5 V/m.Since the field is uniform, we can dispense with the calculus, andreplace dU e = (1/8πk)E 2 dv with U e = (1/8πk)E 2 v. The volumeequals the area multiplied by the gap, so we haveU e = (1/8πk)E 2 (area)(gap)1=8π × 9 × 10 9 N·m 2 /C 2 (1.5 × 105 V/m) 2 (10 −4 m 2 )(10 −5 m)= 1 × 10 −10 Jself-check EShow that the units in the preceding example really do work out to bejoules. ⊲ Answer, p. 926Why k is on the bottom example 15It may also seem strange that the constant k is in the denominatorof the equation dU e = (1/8πk)E 2 dv. The Coulomb constant ktells us how strong electric forces are, so shouldn’t it be on top?584 Chapter 10 Fields