Question

The electric potential of a charge distribution is given by the equation V(x) = 3x2y2 + yz3 - 2z3x, where x, y, z are measured in meters and V is measured in volts. Calculate the magnitude of the electric field vector at the position (x,y,z) = (1.0, 1.0, 1.0)

Answer 1

The magnitude of the electric field is
`|E| = 8.602 \ V/m`

Step-by-step explanation:

From the question we are told that

The electric potential is
`V = 3x^2y^2 + yz^3 - 2z^3x`

Generally electric filed is mathematically represented as

`E = - [(dV )/(dx) i + (dV)/(dy) j + (dV)/(dz) \ k]`

So

`E =- ( [6xy^2 - 2z^3] i + [6x^2y+ z^3]j + [3yz^2 -6xz^2])`

at (x,y,z) = (1.0, 1.0, 1.0)

`E = [6(1)(1)^2 - 2(1)^3] i + [6(1)^2(1)+ (1)^3]j + [6(1)(1)^2 -6(1)(1)^2]`

`E =- ([4] i + [7]j + [-3])`

`E =-4i -7j + 3 k`

The magnitude of the electric field is

`|E| = √((-4)^2 + (-7)^2 + (3^2))`

`|E| = 8.602 \ V/m`