Calculating the electric field from the potential

paralian
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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?

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).

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:
on Phys.org
paralian said:
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?
 
G01 said:
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
 
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?
 
G01 said:
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}<br /> =2.00yz^2<br /> =-64[/tex]

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

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

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

The answer in the back of the book is 150 N/C.
 
paralian said:
[tex]2.00xy \frac{\partial z^2}{\partial z}<br /> =2.00xyz<br /> =-48[/tex]

[tex]\sqrt{64^2 + 96^2 + 48^2}<br /> =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}<br /> =4.00xyz\,.[/tex]

That should give you 150.09. :smile:
 

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