An electric potential is given in volts by

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zhillyz
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Homework Statement



An electric potential is given in volts by;

[itex]\phi (x,y,z) = 20x - 12y^2 +2yz[/itex],

where x,y and z are measured in metres. Find the magnitude of the electric field [itex]E[/itex] at the point(1,1,3).

Homework Equations



[itex]E = -\nabla \phi[/itex]

The Attempt at a Solution



[itex]-\nabla \phi = -1*(\frac{\partial \phi}{\partial x} + \frac{\partial \phi}{\partial y} + \frac{\partial \phi}{\partial z})[/itex]
[itex]= -1*((20)+(24y +2z)+(2y)) = -1*(20+24+6+2) = -52Vm^-1[/itex]


I just wanted to know if I have used the right equation, the process is okay and if there is an occasion I should not use this equation? Thanks!
 
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hi zhillyz! :smile:
zhillyz said:
[itex]E = -\nabla \phi[/itex]

[itex]-\nabla \phi = -1*(\frac{\partial \phi}{\partial x} + \frac{\partial \phi}{\partial y} + \frac{\partial \phi}{\partial z})[/itex]

no, ∇ is a vector

(and the magnitude of (a,b,c) isn't a + b + c, is it? :wink:)​
 
I'm sorry perhaps you could elaborate but I think that the electric field 'E' is equal to negative grad phi, grad being the gradient of a vector as you say. Do you merely mean I should not add the sums together at the end as this would make it scalar? ie,

[itex]20i +30j+2k[/itex]
 
zhillyz said:
… grad being the gradient of a vector as you say …

no, phi is a scalar, grad phi is the gradient of a scalar, it is a vector

what is the magnitude of a vector (a,b,c) ? :smile:
 
[itex]-20i + -30j + -2k = E, |E|= \sqrt {(-20^2) + (-30^2) + (-2k^2)}[/itex]

Better?