# Homework Help: Electric field magnitude problem

1. Oct 30, 2015

### gracy

1. The problem statement, all variables and given/known data
The electric potential existing in space is V(x,y,z)=B(xy+yz+zx).Find the expression for the electric field at point P(1,1,1) and it's magnitude if B=10 S.I unit.

2. Relevant equations
View attachment 91091

3. The solution
I actually do have the solution of problem,but I have many doubts regarding the solution.
So want to list them and have them resolved.The solution is
http://images.thedigilibrary.com/notes/courses/IPROFPHenC18T01ST01.ILLUS.1P.pdf
page no 36 to 38.
For those who(their computers )could not open this pdf and so you don't have to scroll down all the way up to page 36

Query: I will mention one ,by the time it gets resolved I will mention second ...
V=B(xy+yz+zx) why?According to the question V(x,y,z)=B(xy+yz+zx)
Where did (x,y,z) go?

2. Oct 30, 2015

### Staff: Mentor

Are you asking why the potential V doesn't have x,y,z components? Remember that potential is a scalar value, not a vector.

3. Oct 30, 2015

### TSny

V(x,y,z) is notation that indicates that the quantity V is a function of the coordinates x, y, and z. Often, the notation is shortened to just V where it is understood that V depends on (x, y, z). It's similar to what you saw in algebra class where you might have a function y = mx + b which you could also write as y(x) = mx + b where y(x) indicates that y is a function of x.

4. Oct 30, 2015

### SammyS

Staff Emeritus
V(x,y,z) = Bâ‹…(xy+yz+zx) is simply saying that the electric potential, V, is a function of x, y, and z. and at some point, (x,y,z), V is given by the formula indicated.

They merely left off the x, y, and z .

5. Oct 30, 2015

### gracy

When we should use extended form of formula of electric field and when we shouldn't?

6. Oct 30, 2015

### Staff: Mentor

You need the extended form to calculate the components of the electric field, which is a vector, term by term.

7. Oct 30, 2015

### Mister T

$V$ is the value of the function.
$V(x,y,z)$ is the function.

They replaced the function with the value of the function.

Suppose you had $V=f(x,y)=x^2-y$.

$f$ is the function, $V$ is the value of the function. It is common practice to sloppily interchange the function with the value of the function.

In this example let's say $V$ is the electric potential. If you give me the values of $x$ and $y$ I can tell you the value of $V$. For example, if $x=2$ and $y=1$ then $V=3$.

The function $f$ is not the electric potential $V$. It's a recipe that that tells me how to find the value of the electric potential $V$ given the values of $x$ and $y$.

Unfortunately, the authors of physics textbooks take the liberty of interchanging these two things, treating them as though they are the same when they are in fact different. In this case what they are doing is using $V$ as the value of the function, but also telling you that the value of $V$ depends on the values of $x$, $y$, and $z$.

8. Oct 31, 2015

### gracy

9. Oct 31, 2015

### gracy

V=B(xy + yz + zx)
So,
$Ex$=$-dV$/$dx$
=â€“ B(y + z)
I did not understand this part.

Last edited: Oct 31, 2015
10. Oct 31, 2015

### Fredrik

Staff Emeritus
The problem appears to be saying that the variables V,x,y,z can be assigned values that are consistent with V=B(xy+yz+zx), where B is just another real number, not a function. So we have
$$\frac{\partial}{\partial x} V=\frac{\partial}{\partial x} B(xy+yz+zx)=B\frac{\partial}{\partial x}(xy+yz+zx).$$

11. Oct 31, 2015

### gracy

How you came to know that B is just a real number?

12. Oct 31, 2015

### Fredrik

Staff Emeritus
The main reason is that if it isn't, the correct way to start is
$$\frac{\partial}{\partial x} V=\frac{\partial}{\partial x} B(xy+yz+zx)=B'(xy+yz+zx)\frac{\partial}{\partial x}(xy+yz+zx).$$ Also, the problem statement includes the words "if B=10", and the solution includes the words "given B=10". It doesn't look like they're talking about a function there.

13. Oct 31, 2015

### gracy

Could you please tell me while taking derivative with respect to a variable when to take constant out and when to take derivative of constant=0?

14. Oct 31, 2015

### gracy

$Ex$=$dV(x,y,z)$/$dx$
Is it right?

15. Oct 31, 2015

### gracy

In place of
$Ex$=$dV$/$dx$
There should be
$Ex$=$dV(x,y,z)$/$dx$
Right?

16. Oct 31, 2015

### Mister T

It's clear from the context, as Fredrik pointed out. The problem is poorly worded:

A better way to state it:

The electric potential existing in space is V(x,y,z)=B(xy+yz+zx) where B is a constant whose value is 10 in SI units. Find the expression for the electric field at point P(1,1,1) and it's magnitude.

Or better yet ...

The electric potential V=B(xy+yz+zx) where B is a constant whose value is 10 in SI units. Find the expression for the electric field at the point (x,y,z)=(1,1,1) and it's magnitude.

Since this is a worked example, the meaning can be determined from the context of the solution.

17. Oct 31, 2015

### Mister T

They call it the extended form, but I prefer to call it the more general form.

$\vec{E}=-\vec{\nabla}{V}=-\hat{i}\frac{\partial}{\partial x}V-\hat{j}\frac{\partial}{\partial y}V-\hat{k}\frac{\partial}{\partial z}V$, where $\vec{E}=\hat{i}E_x+\hat{j}E_y+\hat{k}E_z$.

If we have a case where $\hat{j}\frac{\partial}{\partial y}V=\hat{k}\frac{\partial}{\partial z}V=0$, in other words $E_y=E_z=0$, and

$E=|E_x|=|-\frac{\partial}{\partial x}V|$.

Last edited: Oct 31, 2015
18. Oct 31, 2015

### Mister T

I'm not sure where you are in your education, Gracy. It's confusing me. This is a Calculus I question whereas the math you're encountering in this thread is usually treated in Calculus III. What course are you taking right now that has prompted you to start this thread?

Look at the rules for taking the derivative of a sum. If you're taking the derivative of a variable plus a constant, you get the derivative of the variable plus the derivative of the constant. The derivative of a constant is always zero.

Look at the rules for taking the derivative of a product. If you're taking the derivative of a constant times a variable, you get the constant times the derivative of the variable.

I think you are not retaining these simple rules, so when you encounter more difficult situations you are getting lost. I recommend that you go back over the rules for taking derivatives and practice them.

19. Oct 31, 2015

### Fredrik

Staff Emeritus
You need to use the product rule (fg)'=f'g+fg'. If f happens to be a constant function, then f'=0 and we can use that to simplifiy the right-hand side to fg'.

20. Oct 31, 2015

### gracy

21. Oct 31, 2015

### Mister T

You have a sign error. See Post #17. Otherwise either of your expressions is valid.

22. Oct 31, 2015

### Fredrik

Staff Emeritus
The notations $\frac{\partial V}{\partial x}$ and $\frac{\partial V(x,y,z)}{\partial x}$ are both acceptable. I would say that the V in the former former expression is the variable that occurs in the equation V=B(xy+yz+zx), and that the V in the latter expression is the function defined implicitly by that equation.

Regardless of notation, only functions have derivatives (variables do not), so the former expression can't be interpreted as a partial derivative of the variable. It must be interpreted as a partial derivative of the function.

You need a minus sign, since the relationship between E and V is $\mathbf E=-\nabla V$.

(I wrote this before I saw that Mister T had already replied).

23. Oct 31, 2015

### Mister T

That's confusing. If we have $y=f(x)$ then $y$ is a variable and $f$ is a function.

We can write $\frac{dy}{dx}=\frac{d}{dx}f(x)$.

It appears to me that on the left-hand side of that equation we have the derivative of a variable. On the right-hand side we have something that we call "the derivative of a function" but I'm not sure if that's a proper way to say it, strictly speaking, because one could argue that it's $\frac{df}{dx}$ that's the derivative of a function. I'm not a mathematician and have only an undergraduate minor in the topic. As a physics major, though, I was exposed to lots more math taught to me by mathematicians, but I spent a lot more time doing math in physics classes, and "learning" it from physicists.

24. Oct 31, 2015

### Fredrik

Staff Emeritus
Right. You could also say that the symbols x,y and f are all variables, and that the difference is that f is assigned values from the set $\mathbb R^\mathbb R$ (the set of functions from $\mathbb R$ into $\mathbb R$), while x and y are assigned values from the set $\mathbb R$.

The derivative of a differentiable function f is defined as the function f' such that
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}$$ for all $x\in\mathbb R$. The derivative of a variable is undefined. It would be weird to define it, since a variable is just a symbol in the language we use to talk about mathematics. When we say "the derivative of f", we're actually talking about the derivative of the thing represented by the variable f. So what would "the derivative of y" be? The derivative of a real number, or the derivative of a letter in the alphabet?

An equation like $y=3x^2$ is a constraint on the values we can assign to x and y. If we choose to assign a value to x, then the constraint assigns a value to y. This means that the constraint implicitly defines a function. So we can define the notation dy/dx even though y isn't a function, by saying that it represents the value at x of the derivative of the function defined by the constraint. If the constraint is y=f(x), then that function is equal to f. So dy/dx should denote the derivative of f, not the derivative of y.

My interpretation of a notation like $\frac{d}{dx}3x^2$ is "the value at x of the derivative of the map $t\mapsto 3t^2$ with domain $\mathbb R$". Any expression of the form $\frac{d}{dx}(\text{some expression involving }x)$ can be interpreted similarly. This intepretation ensures that $\frac{d}{dx}f(x)=f'(x)$. So it also ensures that if our value assignments to variables are constrained by the equation y=f(x), then we have $\frac{d}{dx}y=\frac{d}{dx}f(x)=f'(x)$.

25. Oct 31, 2015

### Mister T

To me $\frac{dy}{dx}$ is read "the derivative of y with respect to x". Saying it "reads" one way, and saying it "denotes" another way would be a pedagogical nightmare. Glad I don't have to teach it!