ON ELECTROSTATIC CAPACITY

Excerpts

ON ELECTROSTATIC CAPACITY
the amount of the charge is the greater the nearer the disks are placed to each other, being approximately inversely as the distance between them.
ON ELECTROSTATIC CAPACITY
In this case the capacity of the inner conductor is almost or altogether independent of everything but the outer conductor. This is the case in the Leyden jar, and in a cable with a copper core surrounded by an insulator the outside of which is protected by a sheathing of iron wires.
ON ELECTROSTATIC CAPACITY
An apparatus consisting of two insulated conductors, each presenting a large surface to the other with a small distance between them, is called a *condenser*, because a small electromotive force is able to charge such an apparatus with a large quantity of electricity.
ON ELECTROSTATIC CAPACITY
If we now remove one of the disks from the other we do work against the electric attraction which draws them together, and we may thus increase the energy of the system so much that, though the original electromotive force was only that of a single voltaic cell, either of the disks when separated may be raised to so high a potential that the gold leaves of an electrometer connected with it are deflected.
ON ELECTROSTATIC CAPACITY
The capacity of a conductor is measured by the charge of electricity which will raise its potential to the value unity, the potential of all other conductors in the field being kept at zero.
ON ELECTROSTATIC CAPACITY
It was in this way that Volta demonstrated that the electrification due to a voltaic cell is of the same kind as that due to friction, the copper electrode being positive with respect to the zinc electrode.
ON ELECTROSTATIC CAPACITY
There are other forms of condensers, however, in which one of the conductors is almost or altogether surrounded by the other. In this case the capacity of the inner conductor is almost or altogether independent of everything but the outer conductor. This is the case in the Leyden jar, and in a cable with a copper core surrounded by an insulator the outside of which is protected by a sheathing of iron wires.
ON ELECTROSTATIC CAPACITY
The following method, by which the existence of a determinate relation between the capacities of four condensers may be verified, has been employed by Sir W. Thomson.% 0.14emGibson and Barclay. It corresponds in electrostatics to Wheatstone’s Bridge in current electricity.
ON ELECTROSTATIC CAPACITY
From this it appears that if we connect first the one and then the other conductor with the earth the values of the potentials and charges will be diminished in the ratio of K^2(K+H)(K+h) to unity.

Equations

ON ELECTROSTATIC CAPACITY
Q &= K(P-p) + HP
The charge of the first conductor is K times its potential difference from the second conductor plus H times its potential, where H is the part of its capacity that depends on external objects.
ON ELECTROSTATIC CAPACITY
q &= K(p-P) + hp
The charge of the second conductor is K times its potential difference from the first conductor plus h times its potential, where h is the part of its capacity depending on external objects.
ON ELECTROSTATIC CAPACITY
P_1=P-\frac{K}{K+H}p
After the second conductor is connected to earth, the potential of the first conductor falls to P_1 by the fraction K/(K+H) of the second conductor's former potential.
ON ELECTROSTATIC CAPACITY
Q_1=(K+H)P_1
After the earthing, the charge of the first conductor equals its capacity K+H times its new potential P_1.
ON ELECTROSTATIC CAPACITY
q_1=-KP_1
After the earthing, the charge of the second conductor equals minus K times the first conductor's new potential.
ON ELECTROSTATIC CAPACITY
p_2=-\frac{K}{K+h}P_1
When the second conductor is insulated and the first is earthed, the second conductor's potential becomes minus K/(K+h) times P_1.
ON ELECTROSTATIC CAPACITY
Q_2=-Kp_2
After the first conductor is earthed, the charge of the first conductor equals minus K times the second conductor's potential p_2.
ON ELECTROSTATIC CAPACITY
q_2=(K+h)p_2
After the first conductor is earthed, the charge of the second conductor equals its capacity K+h times its potential p_2.
ON ELECTROSTATIC CAPACITY
P_3=-\frac{K}{K+H}p_2
When the second conductor is earthed again, the first conductor's potential becomes minus K/(K+H) times p_2.
ON ELECTROSTATIC CAPACITY
Q_3=(K+H)P_3
After the second earthing, the charge of the first conductor equals its capacity K+H times P_3.
ON ELECTROSTATIC CAPACITY
q_3=-KP_3
After the second earthing, the charge of the second conductor equals minus K times the first conductor's potential P_3.
ON ELECTROSTATIC CAPACITY
Q_1=(K_1+H_1)P
The inner coating of the first jar, held at potential P, carries charge (K_1+H_1) times P when its outer coating is at earth.
ON ELECTROSTATIC CAPACITY
Q_2=-K_2 P
The inner coating of the second jar, with its outer coating at earth, carries charge minus K_2 times P.
ON ELECTROSTATIC CAPACITY
Q_1 + Q_2 = {Q_1}' + {Q_2}'
The total charge on the two inner coatings is unchanged when they are joined together.
ON ELECTROSTATIC CAPACITY
{P_1}' = {P_2}' = P'
After the inner coatings are joined, both jars have the same common potential P'.
ON ELECTROSTATIC CAPACITY
(K_1+H_1-K_2)P=(K_1+H_1+K_2+H_2)P'
Charge conservation applied to the joined inner coatings gives the common potential P' in terms of P and the jar capacities.
ON ELECTROSTATIC CAPACITY
K_1 + H_1 = K_2
When the first jar's total capacity equals the second jar's mutual capacity, the discharge between the jars is complete.
ON ELECTROSTATIC CAPACITY
a &= (P+R+\alpha+\eta)A-PB-RD-\eta C
The charge of conductor alpha is its coefficient of capacity times its potential, minus the induction terms from the potentials of the adjacent conductors.
ON ELECTROSTATIC CAPACITY
b &= (P+Q+\beta+\xi)B-PA-QC-\xi D
The charge of conductor beta is its coefficient of capacity times its potential, minus the induction terms from the adjacent conductors.
ON ELECTROSTATIC CAPACITY
c &= (Q+S+\gamma+\eta)C-QB-SD-\eta A
The charge of conductor gamma is its coefficient of capacity times its potential, minus the induction terms from the adjacent conductors.
ON ELECTROSTATIC CAPACITY
d &= (R+S+\delta+\xi)D-RA-SC-\xi B
The charge of conductor delta is its coefficient of capacity times its potential, minus the induction terms from the adjacent conductors.
ON ELECTROSTATIC CAPACITY
a &= (P+R+\alpha+\eta)A-\eta C
Before the discharge, with beta and delta at zero potential, the charge of alpha depends only on A and C.
ON ELECTROSTATIC CAPACITY
b &= \hphantom{(P+R+\alpha}-PA-QC
Before the discharge, with beta at zero potential, the charge of beta depends on A and C through the induction terms.
ON ELECTROSTATIC CAPACITY
c &= (Q+S+\gamma+\eta)C-\eta A
Before the discharge, with beta and delta at zero potential, the charge of gamma depends only on C and A.
ON ELECTROSTATIC CAPACITY
a' + c' = a + c
The total charge of the connected conductors alpha and gamma is the same after the discharge as before.
ON ELECTROSTATIC CAPACITY
b' = b
The charge of conductor beta is unchanged by the discharge between alpha and gamma.
ON ELECTROSTATIC CAPACITY
A' = C' = y
After the discharge, alpha and gamma share the common potential y.
ON ELECTROSTATIC CAPACITY
a' &= (P+R+\alpha)y-PB'
After the discharge, the charge of alpha is expressed through the common potential y and the potential B' of beta.
ON ELECTROSTATIC CAPACITY
b' &= (P+Q+\beta+\xi)B' - (P+Q)y
After the discharge, the charge of beta is expressed through its potential B' and the common potential y.
ON ELECTROSTATIC CAPACITY
c' &= (Q+S+\gamma)y-QB'
After the discharge, the charge of gamma is expressed through the common potential y and the potential B' of beta.
ON ELECTROSTATIC CAPACITY
(P+R+Q+S+\alpha+\gamma)y-(P+Q)B'=(P+R+\alpha)A+(Q+S+\gamma)C
Charge conservation for alpha and gamma after the connection relates the common potential y, the potential B', and the initial potentials A and C.
ON ELECTROSTATIC CAPACITY
(P+Q+\beta+\xi)B'-(P+Q)y = -PA-QC
The unchanged charge of beta, after the discharge, relates its potential B' and the common potential y to the initial potentials A and C.
ON ELECTROSTATIC CAPACITY
B'\{(P+Q)(R+S)+(P+Q)(\alpha+\beta+\gamma+\xi)+(R+S+\alpha+\gamma)(\beta+\xi)\}=\{Q(R+\alpha)-P(S+\gamma)\}(A-C)
Eliminating y gives the potential B' of beta after the discharge in terms of the initial potentials A and C and the coefficients.
ON ELECTROSTATIC CAPACITY
B' = 0
If the electrometer is undisturbed by the discharge, the potential of beta stays zero.
ON ELECTROSTATIC CAPACITY
P : Q :: R + \alpha : S + \gamma
The balance condition: when the electrometer is undisturbed, the capacities of the jars are in the same proportion as the sums of the coefficients on the two arms of the bridge.

Problems

No exercises in this chapter.