Strength of an Electric Field


by boozi
Tags: electric, field, strength
boozi
boozi is offline
#1
Apr29-07, 03:28 PM
P: 8
Hello. This is my first time here, so let me know if I'm doing anything wrong "posting-wise."

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


2. Relevant equations
E = -grad(V)


3. 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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neutrino
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#2
Apr29-07, 03:40 PM
P: 2,048
E = -grad(V) = -(520(x) - 360(y)) = -(-40) = 40
The electric field is a vector field. The potential is a scalar field.
boozi
boozi is offline
#3
Apr29-07, 03:45 PM
P: 8
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?

neutrino
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#4
Apr29-07, 04:02 PM
P: 2,048

Strength of an Electric Field


Given some arbitrary scalar field V = V(x,y), how would you write down its gradient (in Cartesian coordinates)?
boozi
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#5
Apr29-07, 04:05 PM
P: 8
V = partial x + partial y?
Well, grad V = partial x + partial y
neutrino
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#6
Apr29-07, 04:11 PM
P: 2,048
Quote Quote by boozi View Post
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.
boozi
boozi is offline
#7
Apr29-07, 04:13 PM
P: 8
ah ****... i'm stupid..... thx :P


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