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    Question:
    The potential difference between origin and point P(x,y,z) is given by:
    V-V_0 = (30x-15y+xz-4) Volts
    What's the electric field associated (in V/m) at point P?

    My Answer:
    If P has position vector: (x, y, z) then the magnitude of P will be \sqrt{3}.
    The negative standard derivative of the potential vector will be electric field:
    \frac{dV}{dx}=30-15y+z=-30+14\sqrt{3}
    \frac{dV}{dy}=30x-15+xz=12+30\sqrt{3}
    \frac{dV}{dz}=30x-15y+x=-16\sqrt{3}
    Combining the terms:
    E=(\sqrt{3}(14\hat{x}-30\hat{y}-16\hat{z})-18)
    Is this the right answer?
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    (Original post by halpme)
    Question:
    The potential difference between origin and point P(x,y,z) is given by:
    V-V_0 = (30x-15y+xz-4) Volts
    What's the electric field associated (in V/m) at point P?

    My Answer:
    If P has position vector: (x, y, z) then the magnitude of P will be \sqrt{3}.
    The negative standard derivative of the potential vector will be electric field:
    \frac{dV}{dx}=30-15y+z=-30+14\sqrt{3}
    \frac{dV}{dy}=30x-15+xz=12+30\sqrt{3}
    \frac{dV}{dz}=30x-15y+x=-16\sqrt{3}
    Combining the terms:
    E=(\sqrt{3}(14\hat{x}-30\hat{y}-16\hat{z})-18)
    Is this the right answer?
    Firstly, if x, y and z are unknown, how is your magnitude \sqrt{3}. That would be assuming x = y = z = 1.

    The electric field is defined as, E = -\nabla V. As V_0 is constant, \nabla (V-V_0) = \nabla V.

    Now, in your working you used the derivative, which in this case in wrong. You need to be using partials.

    I.e. \dfrac{\partial V}{\partial x} = 30 + z.

    As both y and z are held constant.

    Try have a go now. Remember:

    a(x,y,z)

    \nabla a = \dfrac{\partial a}{\partial x}\hat{i} + \dfrac{\partial a}{\partial y}\hat{j} + \dfrac{\partial a}{\partial z}\hat{k}
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    (Original post by Phichi)
    Firstly, if x, y and z are unknown, how is your magnitude \sqrt{3}. That would be assuming x = y = z = 1.

    The electric field is defined as, E = -\nabla V. As V_0 is constant, \nabla (V-V_0) = \nabla V.

    Now, in your working you used the derivative, which in this case in wrong. You need to be using partials.

    I.e. \dfrac{\partial V}{\partial x} = 30 + z.

    As both y and z are held constant.

    Try have a go now. Remember:

    a(x,y,z)

    \nabla a = \dfrac{\partial a}{\partial x}\hat{i} + \dfrac{\partial a}{\partial y}\hat{j} + \dfrac{\partial a}{\partial z}\hat{k}
    So, having reviewed partial derivatives a little bit:
    I get E=2z+15. Is this all the question asks for?
 
 
 
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