ELECTRIC FIELD. 



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of the cylindrical surface is 2irrl square centimeters, and the 

 amount of electric flux which crosses it is 2wrl X ke. But the 

 amount of charge on / centimeters of the inner cylinder is lq. 

 Therefore, according to Gauss's theorem, we have: 



from which we get: 



JL I ? 



~ 27r k r 



(i) 



For the simplest case, namely when the space between the 

 cylinders is rilled with air, then k = I and this equation becomes : 



2-ir r 



(2) 



Remark. This equation also expresses the intensity e of the 

 electric field at a point r centimeters from the axis of a long 

 cylinder which is at a great distance from all other bodies, q 

 being the charge in coulombs on each centimeter of length of the 

 cylinder. This is evident when we consider that equation (2) 



