80. We can express Eq. 23-17 in terms of the charge density ρas follows: E= q4πεo r2= ρ43πR34πεo r2= ρR33εo r2. Thus, at r= 2R, we have (when the ball is solid) E1= ρR33εo (2R)2= ρ12εo R. Now, with the hollow core of radius R/2, we have a similar field but without the contribution from those charges that would have been in that core:
This is the end of the preview.
access the rest of the document.
Electric charge, Fundamental physics concepts, εo r2