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

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)

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Hưng Khoa 3 years 2021-08-23T15:52:17+00:00 1 Answers 21 views 0

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    2021-08-23T15:53:18+00:00

    Answer:

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

    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 =  - [\frac{dV }{dx} i  +  \frac{dV}{dy} j + \frac{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| =  \sqrt{(-4)^2 + (-7)^2 + (3^2)}

           |E| =  8.602 \ V/m

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