Equipotential Curves and Surfaces, fyc. 



17 



Draw round C as centre the circle which cuts the circumference of 

 A A : Bj B at right angles, and let it cut the circle in H and K, and the 

 chords in E and E x . 



Then CA . CA^CE^CK 3 , 



and CB.CB 1 =CE 2 1 = CH 2 . 



If the value of the constant c is zero in the above expression, then 

 the centre C is a point of zero potential. 



If we produce the other two chords of the circle to meet in the point O, 

 and take a point D in AB, such that OA . OB = OD 2 , and draw a circle 

 about with OD as radius, then this circle is an equipotential curve for 

 the four electrodes A, A v B, B r For if P be any point of this circle, 

 then PB : PA : : DB : DA, 



and PB t : PA X : : D 1 B 1 : D X A X : : DB X : DA,. 



But the potential at any point, P, due to the four electrodes is 

 A , /PA.PAA 



hence on the circle DD X this becomes 



A , /DA.DAA 



A log (dbtdb;) 



which is the same as the potential at the point D. 



Pig. 2. 



Since DA : DB : : OD : OB, 



and D X A X : D^ : : OA x : OD x , or : : OD x : OB L , 



the potential at D is 



These results show that the circle through DD X is an equipotential 

 tol. xxiv. c 



