# Calculating the electric field from the potential

1. Mar 25, 2008

### paralian

1. The problem statement, all variables and given/known data

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?

2. Relevant equations

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

3. 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 $$2.00xyz^2$$.

Last edited: Mar 25, 2008
2. Mar 25, 2008

### G01

It is a function describing V. What is actually confusing you about it? Is it that it involves more than one variable?

3. Mar 26, 2008

### paralian

Haha...yes. It's probably something really simple. I just don't know what to do to find each component

4. Mar 28, 2008

### G01

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?

5. Mar 30, 2008

### paralian

$$2.00yz^2 \frac{\partial x}{\partial x} =2.00yz^2 =-64$$

$$2.00xz^2 \frac{\partial y}{\partial y} =2.00xz^2 =96$$

$$2.00xy \frac{\partial z^2}{\partial z} =2.00xyz =-48$$

$$\sqrt{64^2 + 96^2 + 48^2} =125$$

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

6. Mar 30, 2008

### tiny-tim

Hi paralian!

It should be $$2.00xy \frac{\partial z^2}{\partial z} =4.00xyz\,.$$

That should give you 150.09.