38 



Electrostatics Field of Force 



[CH. II 



a closed conductor is zero, we have experimental proof of the law of the 

 inverse square which admits of much greater accuracy than the experimental 

 proof of Coulomb. 



The theorem that if there is no charge inside a spherical conductor the 

 law of force must be that of the inverse square is due to Laplace. We need 

 consider this converse theorem only in its application to a spherical conductor, 

 this being the actual form of conductor used by Cavendish. The apparatus 

 illustrated in fig. 10 is not that used by Cavendish, but .is an improved 

 form designed by Maxwell, who repeated Cavendish's experiment in a more 

 delicate form. 



Two spherical shells are fixed by a ring of ebonite so as to be concentric 

 with one another, and insulated from one another. 

 Electrical contact can be established between the two 

 by letting down the small trap-door B through which 

 a wire passes, the wire being of such a length as just 

 to establish contact when the trap-door is closed. The 

 experiment is conducted by electrifying the outer 

 shell, opening the trap-door by an insulating thread 

 without discharging the conductor, afterwards dis- 

 charging the outer conductor and testing whether any 

 charge is to be found on the inner shell by placing it 

 in electrical contact with a delicate electroscope by 

 means of a conducting wire inserted through the trap- 

 door. It is found that there are no traces of a charge 

 on the inner sphere. 



F IO> 10. 47. Suppose we start to find the law of electric 



force such that there shall be no charge on the inner 



sphere. Let us assume a law of force such that the repulsion between two 

 charges e, e f at distance r apart is ee'$ (r). The potential, calculated as 

 explained in 33, is 



(13), 



where the summation extends over all the charges in the field. 



Let us calculate the potential at a point inside the sphere due to a charge 

 E spread entirely over the surface of the sphere. If the sphere is of radius a, 

 the area of its surface is 4?ra 2 , so that the amount of charge per unit area is 

 E/4<7ra?, and the expression for the potential becomes 



(14), 



the summation of expression (13) being now replaced by an integration which 

 extends over the whole sphere. In this expression r is the distance from the 



