what is the strength of the electric field at (x,y)=(3.0m, 3.0m)?

where v=(120x^2-290y^2)V, where x and y are in meters.
Please explain. thanks

To find the strength of the electric field at a given point (x, y) in this case, we need to calculate the magnitude of the gradient of the potential function. The formula to calculate the magnitude of the electric field is:

E = -∇V

Where ∇ (pronounced "del") represents the gradient operator. The negative sign comes from the fact that electric field points in the opposite direction of the potential gradient.

In this case, we are given the potential function as V = 120x^2 - 290y^2. To find the electric field, we need to take the partial derivatives of the potential function with respect to x and y. Let's calculate these derivatives:

∂V/∂x = 240x
∂V/∂y = -580y

Now, using these partial derivatives, we can calculate the electric field:

E = -∇V = -(∂V/∂x)i - (∂V/∂y)j

Here, i and j represent the unit vectors in the x and y directions, respectively.

Substituting the partial derivatives we calculated earlier:

E = -(240x)i + (580y)j

The strength of the electric field at a given point (x, y) can be obtained by substituting the values of x and y into the expression obtained above.

In this case, substituting x = 3.0m and y = 3.0m into the expression, we get:

E = -(240 * 3.0)i + (580 * 3.0)j

Simplifying further:

E = -720i + 1740j

Therefore, the strength of the electric field at (x, y) = (3.0m, 3.0m) is -720i + 1740j V/m.