9. Over a certain region of space, the electric potential is V = . (a) Find the expressions for the x, y, and z components of the electric f

Question

9. Over a certain region of space, the electric potential is V = . (a) Find the expressions for the x, y, and z components of the electric field over this region. (b) What is the magnitude of the field at the point P that has coordinates (1.00, 0, 22.00) m?

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Acacia 4 years 2021-09-02T21:39:19+00:00 1 Answers 60 views 0

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  1. thienthanh
    0
    2021-09-02T21:41:17+00:00
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    Answer:

    a)

    The expression for the electric potential in this problem is:

    V=5x-3x^2y+2yz

    where

    x, y, z are the three spatial coordinates

    The relationship between components of the electric field and electric potential is:

    E_x=-\frac{dV}{dx}\\E_y=-\frac{dV}{dy}\\E_z=-\frac{dV}{dz}

    Therefore, we have to calculate the derivatives of the potential over the three variables.

    Doing so, we find:

    E_x=-\frac{d}{dx}(5x-3x^2y+2yz)=-(5-6xy)=6xy-5

    E_y=-\frac{d}{dy}(5x-3x^2y+2yz)=-(-3x^2+2z)=3x^2-2z

    E_z=-\frac{d}{dz}(5x-3x^2y+2yz)=-(2y)=-2y

    b)

    Here we want to find the magnitude of the electric field at the point P that has coordinates

    P (1.00, 0, 22.00) m

    First of all, we find the components of the electric field at that point by substituting

    x = 1.00

    y = 0

    z = 22.0

    We find:

    E_x=6xy-5=6(1)(0)-5=-5 N/C\\E_y=3x^2-2z=3(1)^2-2(22)=-41 N/C\\E_z=-2y=-2(0)=0

    Now, the magnitude of the electric field is given by

    E=\sqrt{E_x^2+E_y^2+E_z^2}

    And by substituting,

    E=\sqrt{(-5)^2+(-41)^2+0}=41.3 N/C

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