Electric field from potential and coordinate

[axgalloway]

Homework Statement

I am given V(x,y,z) = 3x^2 + 2y + 5 and I am given s = (5,3,1), so what is the electric field?

Homework Equations

V= Es

The Attempt at a Solution

I really have no idea. What I tried:

Derivative of V(x,y,z) = 6x+2
so E = 32 N/C?

Really I just don't understand the math necessary. How is it done?

 

[Doc Al]

What you really need to do is take the gradient of the potential function:

[tex]E = - \nabla V[/tex]

You'll find the components of the electric field by taking the partial derivative with respect to each variable. Thus the x-component of the field will be -6x.

See: http://hyperphysics.phy-astr.gsu.edu/Hbase/gradi.html"

 

[nlightner]

I would first clarify with the person giving the homework whether they meant electric field or electric potential. The equation given, V(x,y,z) = 3x^2 + 2y + 5, represents the electric potential at a point in space, not the electric field. The electric field is a vector quantity and is given by the equation E = -∇V, where ∇ is the gradient operator. In this case, the electric field would be given by E(x,y,z) = (6x, 2, 0).

To calculate the electric field at the point (5,3,1), we would substitute those values into the equation to get E(5,3,1) = (30, 2, 0) N/C. It is important to note that the units of electric field are Newtons per Coulomb (N/C), not just Newtons (N).

If the homework truly meant to find the electric field from the given potential, then the attempted solution is correct. However, it is important to clarify and understand the concepts and equations involved in order to solve the problem accurately.