79. (a) At A, the only field contribution is from the +5.00 pC particle in the hollow (this follows from Gauss’ law — it is the only charge enclosed by a Gaussian spherical surface passing through point A, concentric with the shell). Thus, using kfor 1 40πε, we have 122||(5.0010) (0.5)0.180Ek−=×=G. (b) The direction is radially outward. (c) Point
This is the end of the preview.
access the rest of the document.
Electric charge, Gaussian curvature, Torus, Gaussian spherical surface