Take any horizontal line and divide it into small elements of length each representing dq, and draw vertical lines representing the potentials v, v', &c., and after each dose.
If the quadrants of an electrometer are con - nected to the ends of a non-inductive circuit in series with the power-absorbing circuit, and if the needle is connected to the end of this last circuit opposite to that at which the inductionless re - sistance is connected, then the deflexion of the electrometer will be proportional to the power taken up in the circuit, since it is pro - portional to the mean value of (A - B) IC - 1 (A ±B)}, where A and B are the potentials of the quadrants and C is that of the needle.
In any case, therefore, in which we can sum up the elementary potentials at any point we can calculate the resultant electric force at the same point.
Let V 1 and V2 be the potentials of the plates, and let a charge Q be given to one of them.
It depends on the principle that if two condensers of capacity C I and C2 are respectively charged to potentials V I and V2, and then joined in parallel with terminals of opposite charge together, the resulting potential difference of the two condensers will be V, such that V = (C,V 2 -CiV2) /(C1+C2) (16); and hence if V is zero we have C I: C2 = V2 The method is carried out by charging the two condensers to be compared at the two sections of a high resistance joining the ends of a battery which is divided into two parts by a movable contact.'