Register to reply 
Electric Field/Electric Potential (Gradient Notation) 
Share this thread: 
#1
Oct2707, 07:17 PM

P: 171

1. The problem statement, all variables and given/known data
Hey, I have a question about Electric Field/Electric Potential gradient notation. Since, [tex] {\vec{E}} = {}{\nabla}{V(r)} [/tex] Which reduces to, [tex] \vec{E} = {}{\nabla}{V(x, y, z)} [/tex] When expanded is, [tex] \vec{E} = {}{\left[{\frac{\partial[V]}{\partial{x}}}{\hat{i}} + {\frac{\partial[V]}{\partial{y}}}{\hat{j}} + {\frac{\partial[V]}{\partial{z}}}{\hat{k}}\right]} [/tex] So using partial derivative notation can I write, [tex] {\vec{E}} = {}{\vec{V}'_{xyz}} [/tex] So, is the above correct notation? The reason I am hesitant is, because formally the gradient is defined as a vector operator that takes a scalar field (such as the electric potential) and changes it to a vector field (such as the electric field) through: partial differentiation with the addition of unit vectors ([tex]\hat{i}, \hat{j}, \hat{k}[/tex]). However, writing it as below sort of implies the potential is a vector (which it isn't), but gives the impression that it is because of how the gradient is defined. [tex] {\vec{E}} = {}{\vec{V}'_{xyz}} [/tex] So, is the above notation correct? PFStudent 


#2
Oct2707, 07:24 PM

P: 863

[tex] {\vec{E}} = {}{\nabla}{V(r)} [/tex] Or [tex] {\vec{E}} = {}{\nabla}{V} [/tex] 


#3
Oct2807, 10:31 AM

Mentor
P: 8,310

As the above poster says, there is nothing wrong with [itex]\vec{E}=\nabla V[/itex] 


#4
Oct2807, 11:37 AM

P: 171

Electric Field/Electric Potential (Gradient Notation)
Hey,
Yea, thanks for the input, I can see why that notation, [tex] {\vec{E}} = {}{\vec{V}'{xyz}} [/tex] is wrong. Since, we are adding the components of a vector that is not the same as taking the partial derivative of a function with respect to each of the variables. Since, all the gradient is doing is the following, [tex] \vec{E} = {}{\nabla}{V(x, y, z)} = {}{\left[{\frac{\partial}{\partial{x}}{\left[V\right]}} + {\frac{\partial}{\partial{y}}}{\left[V\right]}} + {\frac{\partial}{\partial{z}}}{\left[V\right]}}\right]}{\hat{r}} [/tex] Thanks, PFStudent 


#5
Oct2907, 12:58 PM

P: 171

Hey,
I've been thinking about this and I have a follow up question. Since, [tex] \vec{E} = {}{\nabla}{V(r)} = {}{\left[{\frac{\partial}{\partial{x}}{\left[V\right]}} + {\frac{\partial}{\partial{y}}}{\left[V\right]}} + {\frac{\partial}{\partial{z}}}{\left[V\right]}}\right]}{\hat{r}} [/tex] and also, [tex] E = {}{\frac{\partial}{\partial{r}}}{\left[{V(r)}\right]} [/tex] So then, [tex] \vec{E} = {}{\nabla}{V(r)} = {}{\frac{\partial}{\partial{r}}}{\left[{V(r)}\right]}{\hat{r}} [/tex] Now can I rewrite the above as below? [tex] \vec{E} = {}{\frac{\partial}{\partial{(x, y, z)}}}{\left[{V(x, y, z)}\right]}{\hat{r}} [/tex] Which for [tex]{E}[/tex] can also be written as, [tex] {E} = {}{\frac{\partial}{\partial{(x, y, z)}}}{\left[{V(x, y, z)}\right]} [/tex] So, is the notation for the above two equations correct? Thanks, PFStudent 


Register to reply 
Related Discussions  
Electric potential in relation to electric field problem  Advanced Physics Homework  1  
Calculating the Electric Potential from an Electric Field problem  Introductory Physics Homework  1  
Gradient, potential and electric field  Introductory Physics Homework  0  
Some problems..(electric field.electric potential)  Introductory Physics Homework  3  
Calculating Electric Potential in A Uniform Electric Field  Introductory Physics Homework  4 