Page images
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ав

αγ

By

Ba+Ca-- (Bi2+Ci2) = b. AK, (Bli2+Ck2)−7. BK,

c. AI

βγ
c. Ch1

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c. Al

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1

ay

1

A.

+ a3y [b. BK1. AM + e. CL, ALA

Thus, as far as the sixth term,

. CI,.

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Cmi

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Proceeding in this way, we find that as far as the fourteenth

term,

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1

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By continuing this process we may find any number of

terms of the expression for Ba+Ca.

Now, if E is the charge on A, we have

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Thus the first terms of the expressions for 911 412 413 have been determined and the corresponding terms of 922, 933, 923 may be got by changing the letters. We can get the formula for two spheres from this by making b, AK1, BK1, AL11, BL11 all infinite and C=0.

11

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E=A[a+38
12 = 32 + ...
-B2

-B
...] - B[a +

As an example, suppose we have three small equal conducting spheres, whose centres are at the corners of an equilateral triangle the length of whose side is c. The terms

a

3

we have found will give 911, 9129 13 correctly to (~) °

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2a2 2a3
c2

a 2a3
+
C c2

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The energy of the system, each sphere being supposed to have unit charge, is

1/3 За
+

a

15a3),

C

and the force acting on one of them is

√3
c2

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the force being in the case of point charges.

By the above method the potential due to two charged conducting spheres at an external point can be written down easily. Let the centres of the spheres be A and B, their radii a and b, the distance between the centres c, and the potentials U and V. Let the image of any external point P in B be P1, the image of P, in A, P2, the image of P2 in B, Pa, and so on. Also, let the image of P in A be Q1, the image of Q, in B, Q2, the image of Q2 in A, Qa, and so on.

1

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AP.BQ.BQ3. BQ5. AQ2. AQ6 +

If, now, the image of A in B is I, the image of I, in A, I2, the image of I, in B, Is, and so on, and if the image of B in A is J1, the image of J, in B, J2, the image of J, in A, J3, and so on, it is easy to show from similar triangles that

BP. AP1=c. PI1,

AP.BQ. AQ=c. PI,. AI1,

BP.AP1. BP2. AP3=c. PI3. BI2. AI1,
AP.BQ.AQ2. BQ3. AQ1=c. PI ̧ . AI3 . BI2 . AI1,

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which shows that the potential outside the spheres has the same value as that due to a series of point charges at

A, B, I1, J1, 12, J2, 13, J3, 14, J4, &c., equal respectively to

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c. AJ2. BJ1' c. AI. BI2. AI1' c. BJ. AJ2. BJ'

1

1

&c....,

which are the image charges in the usual way of treating the subject.

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Note. The above paper was written before the one that appeared in the March number of the Philosophical Magazine on the same subject. It is, perhaps, unfortunate that the word "image" has been used for "inverse point." The method has, of course, nothing to do with electrical images.

XLV. Some Two-Dimensional Potential Problems connected with the Circular Arc. By W. G. BICKLEY, B.Sc.*

§ 1.

IN

N this paper a method of dealing with potential problems in two dimensions, depending on the use of functions of a complex variable and of the method of images, is applied to the solution of problems connected with an infinitely long lamina, the section of which is a circular arc. The results obtained are interpreted in terms of electricity and hydrodynamics.

§ 2. The first step in the investigation is the determination of the transformation by which the two sides of the arc in the z-plane become the real axis in the plane of an auxiliary variable (=+n). The arc is taken as that part of the circle z=—te" for which a≤0<a, so that the angle subtended at the centre is 2a. For any point on this circle the ratio (1+2)/(+) is purely real, its value being cot that when lies within the above limits, the values of

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Ө

2'

SO

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are purely real, but become complex when lies outside them. Also, when 2, the expression (1) tends to 0 or o according as the root is taken negatively or positively; and when a, the expression has the values +1.

=

* Communicated by the Author.

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