Calculating the electric field from the potential

In summary, the magnitude of the electric field at the point (3.00\hat{i} - 2.00\hat{j} + 4.00\hat{k})m can be calculated by taking the partial derivatives of the given electric potential function V = 2.00xyz^2 with respect to each variable x, y, and z. The components of the electric field are then determined by using the formula E_{s}=-\delta V/ \delta s (where s represents each component), and the magnitude of the electric field is calculated using the Pythagorean theorem. The correct answer is 150 N/C.
  • #1
paralian
14
0

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].
 
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  • #2
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?
 
  • #3
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
 
  • #4
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?
 
  • #5
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}
=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.
 
  • #6
paralian said:
[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:
 

1. How is the electric field calculated from the potential?

The electric field is calculated by taking the negative gradient of the electric potential. This can be represented mathematically as E = -∇V, where E is the electric field, V is the electric potential, and ∇ is the gradient operator.

2. What is the relationship between electric potential and electric field?

The electric potential is a scalar quantity that represents the potential energy per unit charge at a given point in space. The electric field, on the other hand, is a vector quantity that represents the direction and magnitude of the force on a positive test charge at a given point in space. The electric field is related to the electric potential by the equation E = -∇V, where ∇ is the gradient operator.

3. Can the electric field be negative?

Yes, the electric field can be negative. A negative electric field means that the direction of the force on a positive test charge is opposite to the direction of the electric field. This can occur if the source of the electric field is a negative charge.

4. What are the units of electric field and electric potential?

The units of electric field are newtons per coulomb (N/C) or volts per meter (V/m). The units of electric potential are volts (V) or joules per coulomb (J/C).

5. How does the distance from a charged object affect the electric field and electric potential?

The electric field and electric potential both decrease as distance from a charged object increases. The electric field follows an inverse square law, meaning that it decreases by the square of the distance from the source charge. The electric potential also follows an inverse relationship with distance, meaning that it decreases as distance increases.

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