Calculating the electric field from the potential

AI Thread Summary
To calculate the electric field from the electric potential V = 2.00xyz^2, the components of the electric field are derived using partial derivatives. The confusion arises from understanding how to apply these derivatives while treating the other variables as constants. The calculations for the x, y, and z components yield values of -64, 96, and -48, respectively. However, the correct approach to the z component reveals that it should be 4.00xyz, leading to a total electric field magnitude of approximately 150 N/C, which aligns with the answer in the textbook. This highlights the importance of accurately applying partial derivatives in multivariable functions.
paralian
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Homework Statement



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

Homework Equations



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

The Attempt at a Solution



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

I don't know what they mean by 2.00xyz^2.
 
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paralian said:
I don't know what they mean by 2.00xyz^2.

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:

\frac{\partial V}{\partial x}.

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:

\frac{\partial V}{\partial x}.

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?

2.00yz^2 \frac{\partial x}{\partial x}<br /> =2.00yz^2<br /> =-64

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

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

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

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

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

The answer in the back of the book is 150 N/C.

Hi paralian! :smile:

It should be 2.00xy \frac{\partial z^2}{\partial z}<br /> =4.00xyz\,.

That should give you 150.09. :smile:
 
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