134 



Prof. A. Anderson on Scalar and Vector 



Hence the element contributed to the integral A is 



dq 



V 1 -? 



^sin 2 



where r = PQ is the distance of the actual charge from P 

 and 6 is the angle which the direction of motion makes 

 with QP. 



With the aid of this expression we must now find the 

 value of A for any point P inside a sphere of radius a whose 

 centre is 0, the electricity in it being uniformly distributed 

 and of density p, with a uniform velocity u parallel to the 



axis of z. 



Fie. 3. 



In fig. 3, let OP = b, and let the direction cosines of OP 

 be I, m, n. Also, let axes be drawn through P parallel to 

 the rectangular axes through 0. Draw a cone of small 

 solid angle dco having its vertex at P and intersecting the 

 sphere at L and M. We first find the part of the integral A 

 contributed by this cone. 



This is 



dco 



f- 



p r 2 dr 



n 2 +r« 



dco 



\A-S 



^-sin 2 



c l 



V- 



sin 2 6 



where PL = r 1 , PM = f 2 , and = angle IjPz. Let the plane 



