Saved Bookmarks
| 1. |
Some equipotential surfaces are shown in figure(29.E3) What can you say about the magnitude and the direction of the electric field? |
|
Answer» Solution :` (a) The angle between POTENTIAL edl=dv CHANGE in potential =10V=dV ` ` As E=rdV ` ` (As potential surface) ` ` S0, Edl=dv ` ` rArr Edlcos(90+30) =-dV ` ` rArr E(10xx10^-2)cos120=-dV ` ` E=200V//m ` MAKING an angle 120 with yaxis (b) As ELECTRIC field intensity is rto potential surface ` So, E=r ` ` rArr =60vq ` ` So, E=v.m=V.M. ` |
|