Question

the
electric potential in the region of space near
the point (-2m, 4m, 6m) is V = 80x² +60y^2
what is the electric field vector at that point
​

Answer 1

Answer:

as v=Ed

Explanation:

therfore magnitude of potential at that point will be=80×4m2+60×16m2

=1280m2 volt

disatance of point from origin=✓(x1-x2)2+(y1-y2)2+(z1-z2)2

=d=2m✓14

therefore electric field atthat point is=v/d

E=1280m2/2m✓14

E=320/7×✓14×m

E=171.04×m