• Support PF! Buy your school textbooks, materials and every day products Here!

Calculating the electric field from the potential

  • Thread starter paralian
  • Start date
14
0
1. Homework Statement

What is the magnitude of the electric field at the point (3.00[tex]\hat{i}[/tex] - 2.00[tex]\hat{j}[/tex] + 4.00[tex]\hat{k}[/tex])m if the electric potential is given by V = [tex]2.00xyz^2[/tex], where V is in volts and x, y, and z are in meters?

2. Homework Equations

To calculate the field from the potential, [tex]E_{s}=-\delta V/ \delta s[/tex] (ie [tex]E=-dV/ds[/tex] for each component).

3. The Attempt at a Solution

[tex]E_{x}=-dV_{x}/dx[/tex]
[tex]E_{y}=-dV_{y}/dy[/tex]
[tex]E_{z}=-dV_{z}/dz[/tex]

I don't know what they mean by [tex]2.00xyz^2[/tex].
 
Last edited:

Answers and Replies

G01
Homework Helper
Gold Member
2,649
16
I don't know what they mean by [tex]2.00xyz^2[/tex].
It is a function describing V. What is actually confusing you about it? Is it that it involves more than one variable?
 
14
0
It is a function describing V. What is actually confusing you about it? Is it that it involves more than one variable?
Haha...yes. It's probably something really simple. I just don't know what to do to find each component
 
G01
Homework Helper
Gold Member
2,649
16
O ok. Maybe it is the concept of the partial derivative that is confusing you?

For instance when you want to find the x component of the field you need to find:

[tex]\frac{\partial V}{\partial x}[/tex].

In this case, just think of y and z as constants. Similarly, think of x as constant when taking the derivatives with respect to the other variables.

Does this help?
 
14
0
O ok. Maybe it is the concept of the partial derivative that is confusing you?

For instance when you want to find the x component of the field you need to find:

[tex]\frac{\partial V}{\partial x}[/tex].

In this case, just think of y and z as constants. Similarly, think of x as constant when taking the derivatives with respect to the other variables.

Does this help?
[tex]2.00yz^2 \frac{\partial x}{\partial x}
=2.00yz^2
=-64[/tex]

[tex]2.00xz^2 \frac{\partial y}{\partial y}
=2.00xz^2
=96[/tex]

[tex]2.00xy \frac{\partial z^2}{\partial z}
=2.00xyz
=-48[/tex]

[tex]\sqrt{64^2 + 96^2 + 48^2}
=125[/tex]

The answer in the back of the book is 150 N/C.
 
tiny-tim
Science Advisor
Homework Helper
25,789
249
[tex]2.00xy \frac{\partial z^2}{\partial z}
=2.00xyz
=-48[/tex]

[tex]\sqrt{64^2 + 96^2 + 48^2}
=125[/tex]

The answer in the back of the book is 150 N/C.
Hi paralian! :smile:

It should be [tex]2.00xy \frac{\partial z^2}{\partial z}
=4.00xyz\,.[/tex]

That should give you 150.09. :smile:
 

Related Threads for: Calculating the electric field from the potential

Replies
1
Views
4K
Top