Electric field from potential V= -x*y^2+z

[btpolk]

Homework Statement

Derive the electric field for the potential function V= -x*y^2+z

Homework Equations

V=-∫Eds

The Attempt at a Solution

x*y^2-z=∫Eds

d/ds(x*y^2-z)= E*s

(y^2)dx/ds+(2y*x)dy/ds-dzds=E*s

I'm stuck at this point and I'm not really sure if I'm going in the right direction.

 

[ehild]

The electric field is negative gradient of the potential function. Do you know what the gradient is?

ehild

 

[btpolk]

All I was given is what's in the initial question. Should I just divide the s over and call it a day?

 

[btpolk]

On the second step what if I multiplied both sides by s then divided by s^2 to get E by itself?

 

[ehild]

E is the electric field, and it is a vector, having x, y, z components. You get these components as negative partial derivatives of the potential function.

E_x=-∂V/∂x; E_y=-∂V/∂y; E_z=-∂V/∂z.

ehild

 

[btpolk]

So in vector form would it be <y^2,2xy,1> = E ?

 

[ehild]

There is a minus in front of the z component.

ehild