Calculating Electric Field Strength from Potential in a 2D Region

AI Thread Summary
The electric potential in the given region is expressed as V = (260 x^2 - 180 y^2) V. To find the electric field strength at the point (2.00 m, 3.00 m), the equation E = -grad(V) is used, where grad(V) involves calculating the partial derivatives of V with respect to x and y. The correct gradient is grad V = (520x) i + (-360y) j, leading to the electric field vector. The user expresses confusion about accounting for direction and the distinction between scalar and vector fields. Understanding that the electric field is a vector quantity is crucial for correctly interpreting the results.
boozi
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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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E = -grad(V) = -(520(x) - 360(y)) = -(-40) = 40

The electric field is a vector field. The potential is a scalar field.
 
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?
 
Given some arbitrary scalar field V = V(x,y), how would you write down its gradient (in Cartesian coordinates)?
 
V = partial x + partial y?
Well, grad V = partial x + partial y
 
boozi said:
V = partial x + partial y?
Well, grad V = partial x + partial y

I think you need to read some vector calculus again.

\nabla V = \frac{\partial V}{\partial x}\hat{i} + \frac{\partial V}{\partial y}\hat{j}

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:
ah ****... I'm stupid... thanks :P
 
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