Question

Question 13 what is the magnitude of the electric field at the point (1.90 İ . 2.80 j + 9.00 k ) mif the electric potential is given by V-3.70xyZ2, where Vis in volts and x, y, and z are in meters? Number Units the tolerance is +/-5%

Answer 1

First we need to calculate differential form of electric field in 3D and then to put value to find electric field at given position

E = minus (dv/dx i^^+ dv/dy j^^+ dv/dx k^^) since v = 3.7 xy^x^2. E = minus [3.7 yz^2 i^^+ 3.7 x z^2 j^^+ 7.4 xy z k^^] putting value of x, y, z = E = minus [minus 3.7 times 2.8 times 9^2 i^^+ 3.7 times 1.9 times 9^2 j^^minus 7.4 times 1.9 times 2.8 times 9 k^^] E = 839.16 i^^minus 569.43 j^^+ 354.31 k^^so Vertical-bar E Vertical-bar = squareroot (839.16)^2 minus 569.43 j^^+ 354.31 k^^so Vertical-bar E Vertical-bar = squareroot (839.16)^2 + (569.43)^2 + (354.31)^2 = Vertical-bar E Vertical-bar = 1,074.23 & units is V/m.