Question

What is the magnitude of the electric field at the point (2.80i + 1.75j + 4.20k) m if the electric potential is given by V = 2.00xyz2, where V is in volts and x, y, and z are in meters?

Answer 1

V = 2*x*y*z^2

E = -[-dV/dx i + dV/dy j + dV/dz k]

dV/dx = d(2*x*y*z^2)/dx = 2*y*z^2

at given point (2.80 i + 1.75 j + 4.20 k)

dV/dx = 2*1.75*4.20^2 = 61.74

dV/dy = d(2*x*y*z^2)/dz = 2*x*z^2

at given point (2.80 i + 1.75 j + 4.20 k)

dV/dy = 2*2.80*4.20^2 = 98.78

dV/dz = d(2*x*y*z^2)/dz = 4*x*y*z

at given point (2.80 i + 1.75 j + 4.20 k)

dV/dz = 2*2.80*1.75*4.20 = 41.16

So,

E = -(61.74 i + 98.78 j + 41.16 k)

E = -61.74 i - 98.78 j - 41.16 k

|E| = sqrt (61.74^2 + 98.78^2 + 41.16^2)

|E| = 123.54 V/m