# Calculating the electric field from the potential

1. 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?

2. Homework 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 $$2.00xyz^2$$.

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G01
Homework Helper
Gold Member
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?

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
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?

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

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

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

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

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

tiny-tim
Homework Helper
$$2.00xy \frac{\partial z^2}{\partial z} =2.00xyz =-48$$

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

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

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

That should give you 150.09.