HWK7-1 The circuit with the fastest rate of charging will be the one with the least resistance in the circuit. Thus, C would have the most charge and B would have the least.

Note: For c you must describe what you do and show your data and calculations.

HWK7-2 The fringe field outside the capacitor is Efringe = Qs/(2RAe0), where s is the distance between the plates. So there will be less fringe field per unit charge on the plates to slow down the charging process, thus, circuit #2 would have more charge in the same amount of time .

HWK7-3.

a. The needle on the left will move to the right, (since the current is moving in the direction of the arrow). The top right needle will move 10° to the left, the bottom needle on the right will move 10° right, because current in that branch is moving up at that point.

b. The two bulbs would glow about half as long because with both connected, there will be twice as much crossectional area for the current to go thru, and hence twice as much current initially.

HWK7-4. When the plates are brought closer together, the fringe field will be weakened because it will be affected by the plate of opposite charge more. This will enable more charge to come from the batteries. The current will begin to flow again.

HWK7-5. Electrons begin to move, from the negative to the positive side. The fringe field that pushes the electrons is less with the plastic in place than it would be without it, (since the electric field just outside the plates, due to the polarization inside the plastic, opposes the direction of the net electric field caused by the plates alone just outside the plates) so the discharge is slower than it would be without the plastic.