Saved Bookmarks
| 1. |
Howdoes theelectric flux due toa pointchargeenclosedby asphericalGaussiansurface get aaffected whenitsradiusis increased ? |
|
Answer» Solution :As per Gauss’s theorem, Electric`PHI=q/epsilon_0`where q is the CHARGE enclosed by a closed SURFACE through which flux is `phi` Electric flux does not depend on the radius, hence it REMAINS unaffected. |
|