An electric potential is given in volts by

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Homework Help Overview

The problem involves calculating the electric field from a given electric potential function, which is expressed in terms of three spatial variables. The potential function is defined as φ(x,y,z) = 20x - 12y² + 2yz, and the task is to find the electric field E at a specific point in three-dimensional space.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of the gradient operator to the potential function and the interpretation of the resulting vector. Questions arise regarding the calculation of the magnitude of the electric field and the proper handling of vector components.

Discussion Status

The discussion is ongoing, with participants clarifying the distinction between scalar and vector quantities. Some guidance has been provided regarding the correct interpretation of the gradient and its implications for calculating the electric field's magnitude.

Contextual Notes

There is a focus on ensuring that the mathematical operations respect the vector nature of the gradient and the electric field. Participants are also questioning the accuracy of the calculations presented, particularly in terms of vector addition and magnitude computation.

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?
 
yup! :biggrin:

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

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