# Strength of an Electric Field

by boozi
Tags: electric, field, strength
 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!
P: 2,047
 E = -grad(V) = -(520(x) - 360(y)) = -(-40) = 40
The electric field is a vector field. The potential is a scalar field.
 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?
 P: 2,047 Strength of an Electric Field Given some arbitrary scalar field V = V(x,y), how would you write down its gradient (in Cartesian coordinates)?
 P: 8 V = partial x + partial y? Well, grad V = partial x + partial y
P: 2,047
 Quote by boozi 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.
 P: 8 ah ****... i'm stupid..... thx :P

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