103.] As an example of the method of electric images let us calculate the electric state of two spheres whose radii are a and b respectively, and whose potentials are Pa and Pb , the distance between their centres being c. We shall suppose b to be small compared with c.

Fig. 24.TWO SPHERES.

We may consider the actual electrical effects at any point outside the two spheres as due to a series of electric images. In the first place, since the potential of the sphere A is Pa we must place an image at the centre A with a charge aPa . Similarly at B, the centre of the other sphere, we must place a charge bPb . Each of these images will have an image of the second order in the other sphere. The image of B in the sphere a will be at D, where AD = a2 a , and the charge D = − ⋅ bPb . c c The image of A in the sphere b will be at E, where BE = b2 b , and the charge E = − ⋅ aPa . c c Each of these will have an image of the third order. That of E in a will be at F, where a2 a2 c a2 b AF = = 2 , and F

Pa . AE c − b2 c 2 − b2 That of D in b will be at G, where BG = b2 b2 c ab2 = 2 , and G

Pb . DB c − a2 c 2 − a2 The images of the fourth order will be, of G in a at H, where a2 (c 2 − a2 ) a2 a2 b2

and H

− Pb , AG c(c 2 − a2 − b2 ) c(c 2 − a2 − b2 ) of F in B at I, where AH = BI = b2 (c 2 − b2 ) b2 a 2 b2

and I

− Pa . F B c(c 2 − a2 − b2 ) c(c 2 − a2 − b2 )

We might go on with a series of images for ever, but if b is small compared with c, the images will rapidly become smaller and may be neglected after the fourth order.TWO SPHERES. 96 If we now write a2 b

• &c., c 2 − b2 ab a 2 b2 qab = − − − &c., c c(c 2 − a2 − b2 ) ab2 qbb = b + 2
• &c., c − a2 qaa = a + the whole charge of the sphere a will be Ea = qaa Pa + qab Pb , and that of the sphere b will be Eb = qab Pa + qbb Pb . 104.] From these results we may calculate the potentials of the two spheres when their charges are given, and if we neglect terms involving b3 we find 1 1 Pa = Ea + Eb , a c 1 1 a3 Pb = Ea + − 2 2 E. { b c (c − a2 ) } b c The electric energy of the system is 1 11 2 1 1 1 a3 (Ea Pa + Eb Pb ) = Ea + Ea Eb + − 2 2 Eb2 . 2 { } 2 2a c 2 b c (c − a ) The repulsion, R, between the two spheres is measured by the rate at which the energy diminishes as c increases; therefore, R= a3 (2c 2 − a2 ) E − E . a b c2 { c(c 2 − a2 )2 } EbTWO SPHERES. 97 In order that the force may be repulsive it is necessary that the charges of the spheres should be of the same sign, and a3 (2c 2 − a2 ) Ea must be greater than Eb . c(c 2 − a2 )2

Hence the force is always attractive,

1. When either sphere is uninsulated;
2. When either sphere has no charge;
3. When the spheres are very nearly in contact, if their potentials are different. When the potentials of the two spheres are equal the force is always repul- sive.

105.] To determine the electric force at any point just outside of the surface of a conducting sphere connected with the earth arising from the presence of an electrified point A outside the sphere. The electrical conditions at all points outside the sphere are equivalent, as we have seen, to those due to the point A together with its image at B. If e e is the charge of the point A (Fig. 23), the force due to it at P is AP 2 in the direction AP. Resolving this force in a direction parallel to AC and e along the radius, its components are AC in the direction parallel to AC AP 3 e and CP in the direction CP. The charge of the image of A at B is AP 3 CP CP 1 −e , and the force due to the image at P is e ⋅ in the direction CA CA BP 2 P B. Resolving this force in the same direction as the other, its components are CP CB e ⋅ in a direction parallel to CA, and CA BP 3 CP 2 e in the direction P C. CA ⋅ BP 3 If a is the radius of the sphere and if CA = f = ma and AP = r, thenDENSITY OF INDUCED CHARGE. 98 1 1 a and BP = r; and if e is the charge of the point A, the charge m m 1 of its image at B is − e. m e The force at P due to the charge e at A is 2 in the direction AP. Resolving r this force in the direction of the radius and a direction parallel to AC, its components are CB = e ma ⋅ in the direction AC, and r2 r e a ⋅ in the direction CP. r2 r 1 1 1 m e at B is e or e 2 in the 2 m m BP r direction P B. Resolving this in the same directions as the other force, its components are The force at P due to the image − m BC ema e 2 = 3 in the direction CA, and r BP r m ⋅ CP em2 a e 2 or 3 in the direction P C. r BP r The components in the direction parallel to AC are equal but in opposite directions. The resultant force is therefore in the direction of the radius, which confirms what we have already proved, that the sphere is an equipotential surface to which the resultant force is everywhere normal. The resultant force ea is therefore in the direction P C, and is equal to 3 (m2 − 1) in the direction r P C, that is to say, towards the centre of the sphere. From this we may ascertain the surface density of the electrification at any point of the sphere, for, by Coulomb’s law, if σ is the surface density, 4πσ = R, where R is the resultant force acting outwards.99 DENSITY OF INDUCED CHARGE. Hence, as the resultant force in this case acts inwards, the surface density is everywhere negative, and is σ=− 1 ea 2 (m − 1). 4π r3 Hence the surface density is inversely as the cube of the distance from the inducing point A. 106.] In the case of the two spheres A and B (Fig. 24), whose radii are a and b and potentials Pa and Pb , the distance between their centres being c, we may determine the surface density at any point of the sphere A by considering it as due to the action of a charge aPa at A, together with that due to the pairs of points B, D and E, F &c., the successive pairs of images. Putting r = P B, r1 = P E, r2 = P G, &c., we find σ= 1 1 b {(c 2 − b2 )2 − a2 c 2 } Pa

• 3
• &c. ] 4π [ a r1 a2 c 2 − 1 b 2 b2 c 2 c 2 − a2 − b2 a2 2 Pb (c − a )

−

• &c. 4π [ ar3 r2 3 (c 2 − a2 ) {( c 2 − a2 ) c2 } ] If we call B the inducing body and A the induced body, then we may consider the electrification induced on A as consisting of two parts, one de- pending on the potential of B and the other on its own potential. The part depending on Pb is called by some writers on electricity the in- duced electrification of the first species. When A is not insulated it constitutes the whole electrification, and if Pb is positive it is negative over every part of the surface, but greatest in numerical nature at the point nearest to B. The part depending on Pa is called the induced electrification of the sec- ond species. It can only exist when A is insulated, and it is everywhere of the same sign as Pa . If A is insulated and without charge, then the induced electrifications of the first and second species must be equal and opposite. The surface-density is negative on the side next to B and positive on the sideDENSITY OF INDUCED CHARGE. 100 furthest from B, but though the total quantities of positive and negative elec- trification are equal, the negative electrification is more concentrated than the positive, so that the neutral line which separates the positive from the nega- tive electrification is not the equator of the sphere, but lies nearer to B. The condition that there shall be both positive and negative electrification on the sphere is that the value of σ at the points nearest to B and farthest from B shall have opposite signs. If a and b are small compared with c, we may neglect all the terms of the coefficients of Pa and Pb after the first. The b (c − a) values of r lie between c + a and c − a. Hence, if Pa is between Pb (c + a)2 b (c + a) and Pb , there will be both positive and negative electrification on A, (c − a)2 divided by a neutral line, but if Pa is beyond these limits, the electrification of every part of the surface will be of one kind; negative if Pa is below the lower limit, and positive if it is above the higher limit.