An electric potential is given in volts by

[zhillyz]

Homework Statement

An electric potential is given in volts by;

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

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

Homework Equations

E = -\nabla \phi

The Attempt at a Solution

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


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!

 

[tiny-tim]

hi zhillyz!

E = -\nabla \phi

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

no, ∇ is a vector

(and the magnitude of (a,b,c) isn't a + b + c, is it? )​

 

[zhillyz]

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,

20i +30j+2k

 

[tiny-tim]

… 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) ?

 

[zhillyz]

-20i + -30j + -2k = E, |E|= \sqrt {(-20^2) + (-30^2) + (-2k^2)}

Better?

 

[tiny-tim]

yup!

(i'm assuming that last "k" is a misprint )