Question

The electric potential at points in an xy plane is givenby V = (2.9 V/m²)x² -(2.7V/m²)y². What are(a) the magnitude of the electric field at thepoint (3.4 m, 1.9 m) and (b) the angle that thefield there makes with the positive x direction.

Answer 1

V = (2.9 V/m²)x² -(2.7V/m²)y^{2
Ex at x= 3.4m=dv/dx=2.9*3.4*2=19.72N/C
similarly}
Ey at y= 1.9m=dv/dy=-2.7*1.9*2=-10.26N/C
magnitude of E=√19.72*19.72+10.26*10.26=22.22N/C
direction
the E is making an angle of -27.48degree with +ve X axis.