Calculating Electric Field Strength from Potential in a 2D Region

In summary, The electric potential in a given region of space is V = (260x^2 - 180y^2) V, where x and y are in meters. The strength of the electric field at a point (2.00 m, 3.00 m) is found by taking the gradient of V, which is given by E = -grad(V) = -(520(x) - 360(y)) = -(-40) = 40. This is a vector quantity, so it also has a direction. The correct equation for the gradient is \nabla V = \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}.
  • #1
boozi
8
0
Hello. This is my first time here, so let me know if I'm doing anything wrong "posting-wise."

Homework Statement


The electric potential in a region of space is V= (260 x^2 - 180 y^2) V, where x and y are in meters.

What is the strength of the electric field at (2.00 m, 3.00 m) ?
x = 2.00 m
y = 3.00 m


Homework Equations


E = -grad(V)


The Attempt at a Solution


E = -grad(V) = -(520(x) - 360(y)) = -(-40) = 40

I feel really stupid because it's not the right answer... What am I doing wrong? Thanks in advance!
 
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  • #2
E = -grad(V) = -(520(x) - 360(y)) = -(-40) = 40

The electric field is a vector field. The potential is a scalar field.
 
  • #3
I'm still a bit confused... Well, I have to take into account the direction, too, but... How do I account for it in the equation?
 
  • #4
Given some arbitrary scalar field V = V(x,y), how would you write down its gradient (in Cartesian coordinates)?
 
  • #5
V = partial x + partial y?
Well, grad V = partial x + partial y
 
  • #6
boozi said:
V = partial x + partial y?
Well, grad V = partial x + partial y

I think you need to read some vector calculus again.

[tex]\nabla V = \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}[/tex]

The gradient tells you the direction in which the scalar field, V, is increasing the fastest (at some point). Since a direction is involved it is a vector. But remember, the question asks for the strength of the field, which is the magnitude of the field.
 
Last edited:
  • #7
ah ****... I'm stupid... thanks :P
 

1. What is the strength of an electric field?

The strength of an electric field is a measure of the force that a charged particle experiences when placed in that field. It is typically measured in units of Newtons per Coulomb (N/C).

2. What factors affect the strength of an electric field?

The strength of an electric field is affected by two main factors: the magnitude of the charges creating the field and the distance between those charges. The closer the charges are, the stronger the electric field will be.

3. How is the strength of an electric field calculated?

The strength of an electric field is calculated using the equation E = F/q, where E is the electric field strength, F is the force experienced by a charged particle, and q is the magnitude of the charge on the particle.

4. How does the direction of an electric field affect its strength?

The strength of an electric field is not affected by its direction. However, the direction of the electric field does determine the direction of the force that a charged particle will experience when placed in that field.

5. How does the strength of an electric field vary with distance?

The strength of an electric field decreases as the distance from the source charges increases. This is due to the inverse square law, which states that the strength of an electric field is inversely proportional to the square of the distance from the source charges.

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