Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/424

ance $R,$ and let a current be passed along this series from left to right.

Let us first suppose the plates $B_{0},\,B_{1},\,B_{2},$ each insulated and free from charge. Then the total quantity of electricity on each of the plates $B$ must remain zero, and since the electricity on the plates $A$ is in each case equal and opposite to that of the opposed

Fig. 25.

surface they will not be electrified, and no alteration of the current will be observed.

But let the plates $B$ be all connected together, or let each be connected with the earth. Then, since the potential of $A_{1}$ is positive, while that of the plates $B$ is zero, $A_{1}$ will be positively electrified and $B_{1}$ negatively.

If $P_{1},\,P_{2},$ &c. are the potentials of the plates $A_{1},\,A_{2},$ &c., and $C$ the capacity of each, and if we suppose that a quantity of electricity equal to $Q_{0}$ passes through the wire on the left, $Q_{l}$ through the connexion $R_{1},$ and so on, then the quantity which exists on the plate $A_{1}$ is $Q_{0}-Q_{1},$ and we have

$Q_{0}-Q_{1}=C_{1}P_{1}.$

Similarly

$Q_{1}-Q_{2}=C_{2}P_{2},$

and so on.

But by Ohm's Law we have

${\begin{aligned}P_{1}-P_{2}&=R_{1}{\frac {dQ_{1}}{dt}},\\P_{2}-P_{3}&=R_{2}{\frac {dQ_{2}}{dt}}.\end{aligned}}$

If we suppose the values of $C$ the same for each plate, and those of $R$ the same for each wire, we shall have a series of equations of the form