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Question 15 (F): A parallel plate capacitor (the plates are good conductors) is shown in Figure 8. The plates are separated by d. The capacitor has been charged to a voltage E. Before time t = 0, the battery is disconnected from the capacitor, so that at time t = 0 the electric field in the gap is Ē = -E2/d. At time t = 0, the gap is filled with a poor conductor (conductance o), which is and remains in contact with both of the plates. Show that for t > 0, the charge density (õ, to differentiate this quantity from the conductance o) on the positive plate is given by ot CC õ= d e As an aside, show that 7 x B = 0) in the conductor that fills the gap after t = 0. 7 Eco tttttttt 12- E Lacum لات Set up after tzo ヤftTITオオオオト < Conductor (0) cu Figure 8: Left) Capacitor with parallel plates separated by a distance d.