Assertion (A): Electric field intensity at surface of a uniformly charged spherical shell is `\( E \)`. If shell is punctured at a point then intensity at punctured point becomes `\( E/2 \)`.
Reason (R): Electric field intensity due to a spherical charge distribution can be found out by using Gauss law.
Solution:
The field at a puncture is `\( E/2 \)` due to superposition. Gauss's law helps find the field for symmetric distributions, but it doesn't explain the `\( E/2 \)` effect at the puncture directly. Both (A) and (R) are true, but (R) is not the correct explanation of (A).
Leave a Reply