CHAPTER III. ELECTRICITY
243
351. It cannot because the capacitors would he punctured. The voltages in the capacitors will be Vi = 6,000 V; V2 = 3,000 V and Va = 2,000 V, respectively. Solution. It follows from the fact that the charges across the plates of the capacitors are equal that ViC t = V 2 C2 t V 2 C2 = V 3 Ca and Vt v, + Va = V
+
Solving these equations will give the required answer. 352. C 4n (~~1) ~ 4.2 em, When the bar is shifted the capacitance will not change. Note. In order to solve the problem consider the capacitor with the bar inserted in it as a system of two plane capacitors connected in series. 353. The capacitance will be different. It will be larger in the second case. Solution. In the first case there will only be charges on the inside surface of the larger sphere. In the second case the charges will be on both sides (Fig. 295) and the capaciFig. 295 tance of the entire capacitor must be calculated as the capacitance of a system of two capacitors connected in parallel and with plates A Band
Be.
354. F=QE=2n =
355.
(J
356.
v=
2~
V
Q2 2nQ2 Q2 CV2 s; A=Fd=-S-d= 2C =-2-. 2~ ~ 5 cgs electrostatic units.
V &:p
=4 cgs electrostatic units = 1,200 V.
c:;
Solution. The force acting on the upper plate is P = {see the solution to Problem 354) and the capacitance of the capacitor is S C= 4nd
SV 2 P=&id2
and
V=
... /
II
8:td2P --s
357. In the second case. Solution. In the first case when the plates are drawn apart the potential difference remains constant but the capacitance and,there16*